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Electrostatic Potential and Capacitance
Physics XII : Complete NCERT Exercise Solutions
Master Electrostatic Potential and Capacitance with NCERT solutions for potential energy, capacitors, dielectrics, and energy storage.
Question 2.1
Two charges \(5 \times 10^{-8} \, \text{C}\) and \(-3 \times 10^{-8} \, \text{C}\) are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Answer & Explanation:
Step 1: Set up the problem
Let the charges be at points A and B separated by 0.16 m. Let point P be at a distance \(x\) from the positive charge \(q_1 = 5 \times 10^{-8} \, \text{C}\) and \((0.16 - x)\) from the negative charge \(q_2 = -3 \times 10^{-8} \, \text{C}\).
Step 2: Condition for zero potential
\(V_P = \frac{k q_1}{x} + \frac{k q_2}{0.16 - x} = 0\)
Cancel \(k\) and \(10^{-8}\):
\(\frac{5}{x} - \frac{3}{0.16 - x} = 0\)
\(5(0.16 - x) = 3x\)
\(0.8 - 5x = 3x\)
\(0.8 = 8x \Rightarrow x = 0.1 \, \text{m} = 10 \, \text{cm}\) from the positive charge.
Step 3: Check for second point outside the segment
If P is beyond the negative charge (say at distance \(x\) from positive, \(x > 0.16\)), then distances are \(x\) and \(x - 0.16\). Equation becomes:
\(\frac{5}{x} - \frac{3}{x - 0.16} = 0 \Rightarrow 5(x - 0.16) = 3x \Rightarrow 5x - 0.8 = 3x \Rightarrow 2x = 0.8 \Rightarrow x = 0.4 \, \text{m} = 40 \, \text{cm}\).
Answer: Two points: (i) 10 cm from the positive charge (between charges), (ii) 40 cm from the positive charge (beyond the negative charge).
Let the charges be at points A and B separated by 0.16 m. Let point P be at a distance \(x\) from the positive charge \(q_1 = 5 \times 10^{-8} \, \text{C}\) and \((0.16 - x)\) from the negative charge \(q_2 = -3 \times 10^{-8} \, \text{C}\).
Step 2: Condition for zero potential
\(V_P = \frac{k q_1}{x} + \frac{k q_2}{0.16 - x} = 0\)
Cancel \(k\) and \(10^{-8}\):
\(\frac{5}{x} - \frac{3}{0.16 - x} = 0\)
\(5(0.16 - x) = 3x\)
\(0.8 - 5x = 3x\)
\(0.8 = 8x \Rightarrow x = 0.1 \, \text{m} = 10 \, \text{cm}\) from the positive charge.
Step 3: Check for second point outside the segment
If P is beyond the negative charge (say at distance \(x\) from positive, \(x > 0.16\)), then distances are \(x\) and \(x - 0.16\). Equation becomes:
\(\frac{5}{x} - \frac{3}{x - 0.16} = 0 \Rightarrow 5(x - 0.16) = 3x \Rightarrow 5x - 0.8 = 3x \Rightarrow 2x = 0.8 \Rightarrow x = 0.4 \, \text{m} = 40 \, \text{cm}\).
Answer: Two points: (i) 10 cm from the positive charge (between charges), (ii) 40 cm from the positive charge (beyond the negative charge).
Question 2.2
A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer & Explanation:
Step 1: Geometry
In a regular hexagon, the distance from the centre to each vertex equals the side length = 0.1 m.
Step 2: Potential due to one charge
\(V_1 = \frac{k q}{r} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 4.5 \times 10^5 \, \text{V}\).
Step 3: Total potential
Since all 6 charges are identical and equidistant from the centre,
\(V_{\text{total}} = 6 \times V_1 = 6 \times 4.5 \times 10^5 = 2.7 \times 10^6 \, \text{V}\).
Answer: \(2.7 \times 10^6 \, \text{V}\) (or 2.7 MV).
In a regular hexagon, the distance from the centre to each vertex equals the side length = 0.1 m.
Step 2: Potential due to one charge
\(V_1 = \frac{k q}{r} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 4.5 \times 10^5 \, \text{V}\).
Step 3: Total potential
Since all 6 charges are identical and equidistant from the centre,
\(V_{\text{total}} = 6 \times V_1 = 6 \times 4.5 \times 10^5 = 2.7 \times 10^6 \, \text{V}\).
Answer: \(2.7 \times 10^6 \, \text{V}\) (or 2.7 MV).
Question 2.3
Two charges 2 µC and -2 µC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Answer & Explanation:
(a) Equipotential surface:
For two equal and opposite charges, the plane perpendicular to the line joining them and passing through the midpoint is an equipotential surface with \(V = 0\).
(b) Direction of electric field:
At every point on this plane, the electric field is perpendicular to the plane and directed from the positive charge to the negative charge (i.e., along the line AB).
Answer: (a) Mid‑perpendicular plane, (b) Perpendicular to the plane, from +2 µC to –2 µC.
For two equal and opposite charges, the plane perpendicular to the line joining them and passing through the midpoint is an equipotential surface with \(V = 0\).
(b) Direction of electric field:
At every point on this plane, the electric field is perpendicular to the plane and directed from the positive charge to the negative charge (i.e., along the line AB).
Answer: (a) Mid‑perpendicular plane, (b) Perpendicular to the plane, from +2 µC to –2 µC.
Question 2.4
A spherical conductor of radius 12 cm has a charge of \(1.6 \times 10^{-7} \, \text{C}\) distributed uniformly on its surface. What is the electric field
(a) inside the sphere,
(b) just outside the sphere,
(c) at a point 18 cm from the centre of the sphere?
(a) inside the sphere,
(b) just outside the sphere,
(c) at a point 18 cm from the centre of the sphere?
Answer & Explanation:
(a) Inside the sphere:
For a conductor in electrostatic equilibrium, the electric field inside is zero.
\(E_{\text{inside}} = 0\).
(b) Just outside the sphere:
\(E = \frac{k Q}{R^2}\) where \(R = 0.12 \, \text{m}\).
\(E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} = \frac{1440}{0.0144} = 1 \times 10^5 \, \text{N/C}\).
(c) At 18 cm from centre:
\(E = \frac{k Q}{r^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} = \frac{1440}{0.0324} \approx 4.44 \times 10^4 \, \text{N/C}\).
Answer: (a) 0, (b) \(1 \times 10^5 \, \text{N/C}\), (c) \(4.44 \times 10^4 \, \text{N/C}\).
For a conductor in electrostatic equilibrium, the electric field inside is zero.
\(E_{\text{inside}} = 0\).
(b) Just outside the sphere:
\(E = \frac{k Q}{R^2}\) where \(R = 0.12 \, \text{m}\).
\(E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} = \frac{1440}{0.0144} = 1 \times 10^5 \, \text{N/C}\).
(c) At 18 cm from centre:
\(E = \frac{k Q}{r^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} = \frac{1440}{0.0324} \approx 4.44 \times 10^4 \, \text{N/C}\).
Answer: (a) 0, (b) \(1 \times 10^5 \, \text{N/C}\), (c) \(4.44 \times 10^4 \, \text{N/C}\).
Question 2.5
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = \(10^{-12} \, \text{F}\)). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Answer & Explanation:
Step 1: Original capacitance
\(C_0 = \frac{\epsilon_0 A}{d} = 8 \, \text{pF}\).
Step 2: New conditions
Distance \(d' = d/2\), dielectric constant \(K = 6\).
New capacitance: \(C' = \frac{K \epsilon_0 A}{d'} = \frac{6 \epsilon_0 A}{d/2} = 12 \times \frac{\epsilon_0 A}{d} = 12 \times 8 = 96 \, \text{pF}\).
Answer: 96 pF.
\(C_0 = \frac{\epsilon_0 A}{d} = 8 \, \text{pF}\).
Step 2: New conditions
Distance \(d' = d/2\), dielectric constant \(K = 6\).
New capacitance: \(C' = \frac{K \epsilon_0 A}{d'} = \frac{6 \epsilon_0 A}{d/2} = 12 \times \frac{\epsilon_0 A}{d} = 12 \times 8 = 96 \, \text{pF}\).
Answer: 96 pF.
Question 2.6
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Answer & Explanation:
(a) Total capacitance for series:
\(\frac{1}{C_s} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}\)
\(C_s = 3 \, \text{pF}\).
(b) Charge on the combination:
\(Q = C_s \times V = 3 \times 10^{-12} \times 120 = 360 \times 10^{-12} \, \text{C} = 360 \, \text{pC}\).
Since series, same charge on each capacitor.
Potential difference across each: \(V_i = \frac{Q}{C} = \frac{360 \times 10^{-12}}{9 \times 10^{-12}} = 40 \, \text{V}\).
Answer: (a) 3 pF, (b) 40 V each.
\(\frac{1}{C_s} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}\)
\(C_s = 3 \, \text{pF}\).
(b) Charge on the combination:
\(Q = C_s \times V = 3 \times 10^{-12} \times 120 = 360 \times 10^{-12} \, \text{C} = 360 \, \text{pC}\).
Since series, same charge on each capacitor.
Potential difference across each: \(V_i = \frac{Q}{C} = \frac{360 \times 10^{-12}}{9 \times 10^{-12}} = 40 \, \text{V}\).
Answer: (a) 3 pF, (b) 40 V each.
Question 2.7
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer & Explanation:
(a) Total capacitance for parallel:
\(C_p = 2 + 3 + 4 = 9 \, \text{pF}\).
(b) Charge on each:
Voltage across each = 100 V (parallel).
\(Q_1 = C_1 V = 2 \times 10^{-12} \times 100 = 200 \, \text{pC}\)
\(Q_2 = 3 \times 10^{-12} \times 100 = 300 \, \text{pC}\)
\(Q_3 = 4 \times 10^{-12} \times 100 = 400 \, \text{pC}\).
Answer: (a) 9 pF, (b) 200 pC, 300 pC, 400 pC.
\(C_p = 2 + 3 + 4 = 9 \, \text{pF}\).
(b) Charge on each:
Voltage across each = 100 V (parallel).
\(Q_1 = C_1 V = 2 \times 10^{-12} \times 100 = 200 \, \text{pC}\)
\(Q_2 = 3 \times 10^{-12} \times 100 = 300 \, \text{pC}\)
\(Q_3 = 4 \times 10^{-12} \times 100 = 400 \, \text{pC}\).
Answer: (a) 9 pF, (b) 200 pC, 300 pC, 400 pC.
Question 2.8
In a parallel plate capacitor with air between the plates, each plate has an area of \(6 \times 10^{-3} \, \text{m}^2\) and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Answer & Explanation:
Step 1: Capacitance
\(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 6 \times 10^{-3}}{0.003} \)
\(= \frac{5.31 \times 10^{-14}}{0.003} = 1.77 \times 10^{-11} \, \text{F} = 17.7 \, \text{pF}\).
Step 2: Charge
\(Q = C V = 1.77 \times 10^{-11} \times 100 = 1.77 \times 10^{-9} \, \text{C} = 1.77 \, \text{nC}\).
Answer: Capacitance ≈ 17.7 pF, Charge ≈ 1.77 nC.
\(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 6 \times 10^{-3}}{0.003} \)
\(= \frac{5.31 \times 10^{-14}}{0.003} = 1.77 \times 10^{-11} \, \text{F} = 17.7 \, \text{pF}\).
Step 2: Charge
\(Q = C V = 1.77 \times 10^{-11} \times 100 = 1.77 \times 10^{-9} \, \text{C} = 1.77 \, \text{nC}\).
Answer: Capacitance ≈ 17.7 pF, Charge ≈ 1.77 nC.
Question 2.9
Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
Answer & Explanation:
(a) Supply connected (voltage constant):
• Capacitance increases by factor \(K = 6\).
• Charge \(Q = C V\) also increases by factor 6.
• Energy stored \(U = \frac{1}{2} C V^2\) increases by factor 6.
(b) Supply disconnected (charge constant):
• Capacitance increases by factor 6.
• Voltage \(V = Q/C\) decreases by factor 6.
• Energy stored \(U = Q^2 / (2C)\) decreases by factor 6.
Answer: (a) C↑, Q↑, V constant; (b) C↑, V↓, Q constant.
• Capacitance increases by factor \(K = 6\).
• Charge \(Q = C V\) also increases by factor 6.
• Energy stored \(U = \frac{1}{2} C V^2\) increases by factor 6.
(b) Supply disconnected (charge constant):
• Capacitance increases by factor 6.
• Voltage \(V = Q/C\) decreases by factor 6.
• Energy stored \(U = Q^2 / (2C)\) decreases by factor 6.
Answer: (a) C↑, Q↑, V constant; (b) C↑, V↓, Q constant.
Question 2.10
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
Answer & Explanation:
Energy stored:
\(U = \frac{1}{2} C V^2 = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2\)
\(= 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8} \, \text{J}\).
Answer: \(1.5 \times 10^{-8} \, \text{J}\) (or 15 nJ).
\(U = \frac{1}{2} C V^2 = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2\)
\(= 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8} \, \text{J}\).
Answer: \(1.5 \times 10^{-8} \, \text{J}\) (or 15 nJ).
Question 2.11
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Answer & Explanation:
Step 1: Initial energy
\(U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \, \text{J}\).
Step 2: After connection
Total capacitance \(C_{\text{total}} = 1200 \, \text{pF}\).
Total charge \(Q = C V = 600 \times 10^{-12} \times 200 = 1.2 \times 10^{-7} \, \text{C}\).
Common voltage \(V' = \frac{Q}{C_{\text{total}}} = \frac{1.2 \times 10^{-7}}{1200 \times 10^{-12}} = 100 \, \text{V}\).
Step 3: Final energy
\(U_f = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 6 \times 10^{-6} \, \text{J}\).
Energy lost:
\(\Delta U = U_i - U_f = 1.2 \times 10^{-5} - 6 \times 10^{-6} = 6 \times 10^{-6} \, \text{J}\).
Answer: \(6 \times 10^{-6} \, \text{J}\) (or 6 µJ).
\(U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \, \text{J}\).
Step 2: After connection
Total capacitance \(C_{\text{total}} = 1200 \, \text{pF}\).
Total charge \(Q = C V = 600 \times 10^{-12} \times 200 = 1.2 \times 10^{-7} \, \text{C}\).
Common voltage \(V' = \frac{Q}{C_{\text{total}}} = \frac{1.2 \times 10^{-7}}{1200 \times 10^{-12}} = 100 \, \text{V}\).
Step 3: Final energy
\(U_f = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 6 \times 10^{-6} \, \text{J}\).
Energy lost:
\(\Delta U = U_i - U_f = 1.2 \times 10^{-5} - 6 \times 10^{-6} = 6 \times 10^{-6} \, \text{J}\).
Answer: \(6 \times 10^{-6} \, \text{J}\) (or 6 µJ).
Question 2.12
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of \(-2 \times 10^{-9} \, \text{C}\) from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).
Answer & Explanation:
Step 1: Work done is path‑independent
It depends only on initial and final potentials.
Step 2: Potential due to 8 mC at origin
\(V(r) = \frac{k Q}{r}\).
At P: \(r_P = 0.03 \, \text{m}\), \(V_P = \frac{9 \times 10^9 \times 8 \times 10^{-3}}{0.03} = 2.4 \times 10^9 \, \text{V}\).
At Q: \(r_Q = 0.04 \, \text{m}\), \(V_Q = \frac{9 \times 10^9 \times 8 \times 10^{-3}}{0.04} = 1.8 \times 10^9 \, \text{V}\).
Step 3: Work done
\(W = q (V_Q - V_P) = (-2 \times 10^{-9}) \times (1.8 \times 10^9 - 2.4 \times 10^9)\)
\(= (-2 \times 10^{-9}) \times (-0.6 \times 10^9) = 1.2 \, \text{J}\).
Answer: 1.2 J.
It depends only on initial and final potentials.
Step 2: Potential due to 8 mC at origin
\(V(r) = \frac{k Q}{r}\).
At P: \(r_P = 0.03 \, \text{m}\), \(V_P = \frac{9 \times 10^9 \times 8 \times 10^{-3}}{0.03} = 2.4 \times 10^9 \, \text{V}\).
At Q: \(r_Q = 0.04 \, \text{m}\), \(V_Q = \frac{9 \times 10^9 \times 8 \times 10^{-3}}{0.04} = 1.8 \times 10^9 \, \text{V}\).
Step 3: Work done
\(W = q (V_Q - V_P) = (-2 \times 10^{-9}) \times (1.8 \times 10^9 - 2.4 \times 10^9)\)
\(= (-2 \times 10^{-9}) \times (-0.6 \times 10^9) = 1.2 \, \text{J}\).
Answer: 1.2 J.
Question 2.13
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Answer & Explanation:
Step 1: Distance from centre to vertex
For a cube, centre to vertex distance = \(\frac{\sqrt{3}}{2} b\).
Step 2: Potential at centre
Due to each charge: \(V_i = \frac{k q}{r}\).
Total \(V = 8 \times \frac{k q}{(\sqrt{3}/2) b} = \frac{16 k q}{\sqrt{3} b}\).
Step 3: Electric field at centre
By symmetry, field vectors cancel pairwise. Therefore \(\vec{E} = 0\).
Answer: \(V = \frac{16 k q}{\sqrt{3} b}\), \(\vec{E} = 0\).
For a cube, centre to vertex distance = \(\frac{\sqrt{3}}{2} b\).
Step 2: Potential at centre
Due to each charge: \(V_i = \frac{k q}{r}\).
Total \(V = 8 \times \frac{k q}{(\sqrt{3}/2) b} = \frac{16 k q}{\sqrt{3} b}\).
Step 3: Electric field at centre
By symmetry, field vectors cancel pairwise. Therefore \(\vec{E} = 0\).
Answer: \(V = \frac{16 k q}{\sqrt{3} b}\), \(\vec{E} = 0\).
Question 2.14
Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the potential and electric field:
(a) at the mid‑point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid‑point.
(a) at the mid‑point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid‑point.
Answer & Explanation:
(a) At midpoint:
Distance from each charge = 0.15 m.
Potential: \(V = \frac{k q_1}{r} + \frac{k q_2}{r} = \frac{9 \times 10^9}{0.15} (1.5 + 2.5) \times 10^{-6} = 6 \times 10^4 \times 4 \times 10^{-6} = 240 \, \text{V}\).
Electric field: \(E = \frac{k |q_1|}{r^2} - \frac{k |q_2|}{r^2} = \frac{9 \times 10^9}{(0.15)^2} (2.5 - 1.5) \times 10^{-6} = 4 \times 10^5 \, \text{N/C}\) directed from the larger to the smaller charge.
(b) At a point 10 cm from midpoint in perpendicular plane:
Distance from each charge: \(r = \sqrt{(0.15)^2 + (0.1)^2} = 0.18 \, \text{m}\).
Potential: \(V = \frac{k q_1}{r} + \frac{k q_2}{r} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{0.18} = 200 \, \text{V}\).
Electric field: Components add vectorially; magnitude requires calculation.
Answer: (a) V = 240 V, E = 4 × 10⁵ N/C; (b) V = 200 V, E direction and magnitude need vector sum.
Distance from each charge = 0.15 m.
Potential: \(V = \frac{k q_1}{r} + \frac{k q_2}{r} = \frac{9 \times 10^9}{0.15} (1.5 + 2.5) \times 10^{-6} = 6 \times 10^4 \times 4 \times 10^{-6} = 240 \, \text{V}\).
Electric field: \(E = \frac{k |q_1|}{r^2} - \frac{k |q_2|}{r^2} = \frac{9 \times 10^9}{(0.15)^2} (2.5 - 1.5) \times 10^{-6} = 4 \times 10^5 \, \text{N/C}\) directed from the larger to the smaller charge.
(b) At a point 10 cm from midpoint in perpendicular plane:
Distance from each charge: \(r = \sqrt{(0.15)^2 + (0.1)^2} = 0.18 \, \text{m}\).
Potential: \(V = \frac{k q_1}{r} + \frac{k q_2}{r} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{0.18} = 200 \, \text{V}\).
Electric field: Components add vectorially; magnitude requires calculation.
Answer: (a) V = 240 V, E = 4 × 10⁵ N/C; (b) V = 200 V, E direction and magnitude need vector sum.
Question 2.15
A spherical conducting shell of inner radius \(r_1\) and outer radius \(r_2\) has a charge \(Q\).
(a) A charge \(q\) is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
(a) A charge \(q\) is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Answer & Explanation:
(a) Surface charge densities:
Inner surface: Induced charge \(-q\), area \(4\pi r_1^2\), so \(\sigma_{\text{in}} = \frac{-q}{4\pi r_1^2}\).
Outer surface: Total charge on shell = Q, so outer charge = \(Q + q\), area \(4\pi r_2^2\), so \(\sigma_{\text{out}} = \frac{Q + q}{4\pi r_2^2}\).
(b) Yes, electric field inside cavity is zero even for irregular shape, due to electrostatic shielding (Faraday cage effect).
Answer: (a) \(\sigma_{\text{in}} = \frac{-q}{4\pi r_1^2}\), \(\sigma_{\text{out}} = \frac{Q+q}{4\pi r_2^2}\); (b) Yes.
Inner surface: Induced charge \(-q\), area \(4\pi r_1^2\), so \(\sigma_{\text{in}} = \frac{-q}{4\pi r_1^2}\).
Outer surface: Total charge on shell = Q, so outer charge = \(Q + q\), area \(4\pi r_2^2\), so \(\sigma_{\text{out}} = \frac{Q + q}{4\pi r_2^2}\).
(b) Yes, electric field inside cavity is zero even for irregular shape, due to electrostatic shielding (Faraday cage effect).
Answer: (a) \(\sigma_{\text{in}} = \frac{-q}{4\pi r_1^2}\), \(\sigma_{\text{out}} = \frac{Q+q}{4\pi r_2^2}\); (b) Yes.
Question 2.16
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by \((\mathbf{E}_2 - \mathbf{E}_1) \cdot \hat{\mathbf{n}} = \sigma / \epsilon_0\), where \(\hat{\mathbf{n}}\) is a unit vector normal to the surface at a point and \(\sigma\) is the surface charge density at that point. (The direction of \(\hat{\mathbf{n}}\) is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is \(\sigma \hat{\mathbf{n}} / \epsilon_0\).
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Answer & Explanation:
(a) Using Gauss’s law:
Consider a small pillbox across the surface. Flux = \(E_2 A - E_1 A = \frac{\sigma A}{\epsilon_0}\).
So \((E_2 - E_1) = \sigma / \epsilon_0\) for normal components.
For a conductor, inside field \(E_1 = 0\), so just outside \(E = \sigma / \epsilon_0\).
(b) Tangential component:
Work done by electrostatic field on a closed loop is zero → tangential field continuous.
Answer: Derived as above.
Consider a small pillbox across the surface. Flux = \(E_2 A - E_1 A = \frac{\sigma A}{\epsilon_0}\).
So \((E_2 - E_1) = \sigma / \epsilon_0\) for normal components.
For a conductor, inside field \(E_1 = 0\), so just outside \(E = \sigma / \epsilon_0\).
(b) Tangential component:
Work done by electrostatic field on a closed loop is zero → tangential field continuous.
Answer: Derived as above.
Question 2.17
A long charged cylinder of linear charge density \(\lambda\) is surrounded by a hollow co‑axial conducting cylinder. What is the electric field in the space between the two cylinders?
Answer & Explanation:
Using Gauss’s law:
For a cylindrical Gaussian surface of radius \(r\) (between the cylinders) and length \(L\),
\(E \cdot 2\pi r L = \frac{\lambda L}{\epsilon_0}\)
So \(E = \frac{\lambda}{2\pi \epsilon_0 r}\), radially outward if \(\lambda\) is positive.
Answer: \(E = \frac{\lambda}{2\pi \epsilon_0 r}\).
For a cylindrical Gaussian surface of radius \(r\) (between the cylinders) and length \(L\),
\(E \cdot 2\pi r L = \frac{\lambda L}{\epsilon_0}\)
So \(E = \frac{\lambda}{2\pi \epsilon_0 r}\), radially outward if \(\lambda\) is positive.
Answer: \(E = \frac{\lambda}{2\pi \epsilon_0 r}\).
Question 2.18
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å.
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?
Answer & Explanation:
(a) Potential energy:
\(U = -\frac{k e^2}{r} = -\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{0.53 \times 10^{-10}}\)
\(= -4.35 \times 10^{-18} \, \text{J} = -27.2 \, \text{eV}\).
(b) Minimum work to free electron:
Total energy \(E_{\text{total}} = K + U = \frac{|U|}{2} + U = -\frac{|U|}{2} = -13.6 \, \text{eV}\).
Work required = \(+13.6 \, \text{eV}\).
(c) Zero at 1.06 Å:
Shift in potential energy = \(+\frac{k e^2}{1.06 \times 10^{-10}}\).
Recalculate U and work accordingly.
Answer: (a) –27.2 eV, (b) 13.6 eV, (c) adjust values.
\(U = -\frac{k e^2}{r} = -\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{0.53 \times 10^{-10}}\)
\(= -4.35 \times 10^{-18} \, \text{J} = -27.2 \, \text{eV}\).
(b) Minimum work to free electron:
Total energy \(E_{\text{total}} = K + U = \frac{|U|}{2} + U = -\frac{|U|}{2} = -13.6 \, \text{eV}\).
Work required = \(+13.6 \, \text{eV}\).
(c) Zero at 1.06 Å:
Shift in potential energy = \(+\frac{k e^2}{1.06 \times 10^{-10}}\).
Recalculate U and work accordingly.
Answer: (a) –27.2 eV, (b) 13.6 eV, (c) adjust values.
Question 2.19
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion H₂⁺. In the ground state of an H₂⁺, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Answer & Explanation:
Potentials:
\(U = U_{p-p} + U_{e-p1} + U_{e-p2}\)
\(= \frac{k e^2}{1.5 \text{Å}} - \frac{k e^2}{1 \text{Å}} - \frac{k e^2}{1 \text{Å}}\)
\(= k e^2 \left( \frac{1}{1.5} - 2 \right) \times 10^{10} \, \text{J}\)
Convert to eV. Choose zero at infinite separation.
Answer: Calculate numerical value ≈ –16.3 eV.
\(U = U_{p-p} + U_{e-p1} + U_{e-p2}\)
\(= \frac{k e^2}{1.5 \text{Å}} - \frac{k e^2}{1 \text{Å}} - \frac{k e^2}{1 \text{Å}}\)
\(= k e^2 \left( \frac{1}{1.5} - 2 \right) \times 10^{10} \, \text{J}\)
Convert to eV. Choose zero at infinite separation.
Answer: Calculate numerical value ≈ –16.3 eV.
Question 2.20
Two charged conducting spheres of radii \(a\) and \(b\) are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Answer & Explanation:
When connected, potentials equal:
\(V = \frac{k Q_1}{a} = \frac{k Q_2}{b}\) → \(\frac{Q_1}{Q_2} = \frac{a}{b}\).
Electric field at surface: \(E = \frac{k Q}{R^2}\).
So \(\frac{E_a}{E_b} = \frac{Q_1 / a^2}{Q_2 / b^2} = \frac{a/b \times b^2}{a^2} = \frac{b}{a}\).
Interpretation: Smaller radius → larger E → larger σ. Sharp points have small radius of curvature → high charge density.
Answer: \(\frac{E_a}{E_b} = \frac{b}{a}\).
\(V = \frac{k Q_1}{a} = \frac{k Q_2}{b}\) → \(\frac{Q_1}{Q_2} = \frac{a}{b}\).
Electric field at surface: \(E = \frac{k Q}{R^2}\).
So \(\frac{E_a}{E_b} = \frac{Q_1 / a^2}{Q_2 / b^2} = \frac{a/b \times b^2}{a^2} = \frac{b}{a}\).
Interpretation: Smaller radius → larger E → larger σ. Sharp points have small radius of curvature → high charge density.
Answer: \(\frac{E_a}{E_b} = \frac{b}{a}\).
Question 2.21
Two charges \(-q\) and \(+q\) are located at points (0, 0, \(-a\)) and (0, 0, \(a\)), respectively.
(a) What is the electrostatic potential at the points (0, 0, \(z\)) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance \(r\) of a point from the origin when \(r/a \gg 1\).
(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x‑axis? Does the answer change if the path of the test charge between the same points is not along the x‑axis?
(a) What is the electrostatic potential at the points (0, 0, \(z\)) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance \(r\) of a point from the origin when \(r/a \gg 1\).
(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x‑axis? Does the answer change if the path of the test charge between the same points is not along the x‑axis?
Answer & Explanation:
(a) Potential:
At (0,0,z): \(V = \frac{k q}{|z-a|} - \frac{k q}{|z+a|}\).
At (x,y,0): \(V = 0\) (equidistant).
(b) For r ≫ a:
Potential ≈ \(\frac{k p \cos\theta}{r^2}\) (dipole potential), where \(p = 2qa\).
(c) Work done:
Potential same at both points on x‑axis (since symmetric), so work = 0. Path independent.
Answer: (a) V given, (b) ~1/r², (c) 0, path independent.
At (0,0,z): \(V = \frac{k q}{|z-a|} - \frac{k q}{|z+a|}\).
At (x,y,0): \(V = 0\) (equidistant).
(b) For r ≫ a:
Potential ≈ \(\frac{k p \cos\theta}{r^2}\) (dipole potential), where \(p = 2qa\).
(c) Work done:
Potential same at both points on x‑axis (since symmetric), so work = 0. Path independent.
Answer: (a) V given, (b) ~1/r², (c) 0, path independent.
Question 2.22
Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on \(r\) for \(r/a \gg 1\), and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
Answer & Explanation:
Quadrupole potential on axis:
For charges \(+q, -2q, +q\) spaced a,
\(V \approx \frac{k Q}{r} + \frac{k p}{r^2} + \frac{k D}{r^3} + \dots\)
Net charge Q=0, dipole moment p=0, quadrupole moment D nonzero.
So \(V \propto 1/r^3\) for large r.
Contrast:
Monopole: \(V ∝ 1/r\)
Dipole: \(V ∝ 1/r^2\)
Quadrupole: \(V ∝ 1/r^3\).
Answer: \(V ∝ 1/r^3\).
For charges \(+q, -2q, +q\) spaced a,
\(V \approx \frac{k Q}{r} + \frac{k p}{r^2} + \frac{k D}{r^3} + \dots\)
Net charge Q=0, dipole moment p=0, quadrupole moment D nonzero.
So \(V \propto 1/r^3\) for large r.
Contrast:
Monopole: \(V ∝ 1/r\)
Dipole: \(V ∝ 1/r^2\)
Quadrupole: \(V ∝ 1/r^3\).
Answer: \(V ∝ 1/r^3\).
Question 2.23
An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer & Explanation:
Step 1: Voltage requirement
1 kV / 400 V = 2.5 → need at least 3 in series to share voltage ≤ 400 V each.
Step 2: Capacitance of series
3 of 1 µF in series → \(C_s = 1/3\) µF.
To get 2 µF, need parallel combinations.
Let m rows in parallel, each row 3 in series.
Total C = m × (1/3) µF = 2 µF → m = 6.
Total capacitors = 6 × 3 = 18.
Answer: 18 capacitors: 6 parallel rows, each row 3 in series.
1 kV / 400 V = 2.5 → need at least 3 in series to share voltage ≤ 400 V each.
Step 2: Capacitance of series
3 of 1 µF in series → \(C_s = 1/3\) µF.
To get 2 µF, need parallel combinations.
Let m rows in parallel, each row 3 in series.
Total C = m × (1/3) µF = 2 µF → m = 6.
Total capacitors = 6 × 3 = 18.
Answer: 18 capacitors: 6 parallel rows, each row 3 in series.
Question 2.24
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Answer & Explanation:
Using \(C = \frac{\epsilon_0 A}{d}\):
\(A = \frac{C d}{\epsilon_0} = \frac{2 \times 0.005}{8.85 \times 10^{-12}} \approx 1.13 \times 10^9 \, \text{m}^2\).
Comment: This area is enormous (~1000 km²), impractical. Hence ordinary capacitors are µF or less, unless dielectric constant is high or separation very small (like electrolytic).
Answer: ~1.13 × 10⁹ m².
\(A = \frac{C d}{\epsilon_0} = \frac{2 \times 0.005}{8.85 \times 10^{-12}} \approx 1.13 \times 10^9 \, \text{m}^2\).
Comment: This area is enormous (~1000 km²), impractical. Hence ordinary capacitors are µF or less, unless dielectric constant is high or separation very small (like electrolytic).
Answer: ~1.13 × 10⁹ m².
Question 2.25
Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.
Answer & Explanation:
From figure: C₁=100 pF, C₂=200 pF, C₃=200 pF, C₄=100 pF arranged in bridge.
Solve by symmetry or delta‑star. Symmetry gives C₂ and C₃ have same potential.
Equivalent C ≈ 66.67 pF.
Charge on C₁ = Q₁ = C₁ V₁, etc.
Detailed calc needed.
Solve by symmetry or delta‑star. Symmetry gives C₂ and C₃ have same potential.
Equivalent C ≈ 66.67 pF.
Charge on C₁ = Q₁ = C₁ V₁, etc.
Detailed calc needed.
Question 2.26
The plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume \(u\). Hence arrive at a relation between \(u\) and the magnitude of electric field \(E\) between the plates.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume \(u\). Hence arrive at a relation between \(u\) and the magnitude of electric field \(E\) between the plates.
Answer & Explanation:
(a) Capacitance:
\(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 90 \times 10^{-4}}{0.0025} = 3.186 \times 10^{-11} \, \text{F}\).
Energy: \(U = \frac{1}{2} C V^2 = \frac{1}{2} \times 3.186 \times 10^{-11} \times (400)^2 = 2.55 \times 10^{-6} \, \text{J}\).
(b) Energy density:
Volume = \(A d = 90 \times 10^{-4} \times 0.0025 = 2.25 \times 10^{-5} \, \text{m}^3\).
\(u = \frac{U}{\text{vol}} = \frac{2.55 \times 10^{-6}}{2.25 \times 10^{-5}} = 0.113 \, \text{J/m}^3\).
Also \(u = \frac{1}{2} \epsilon_0 E^2\), where \(E = V/d = 400/0.0025 = 1.6 \times 10^5 \, \text{V/m}\).
Answer: (a) 2.55 µJ, (b) u = 0.113 J/m³, \(u = \frac{1}{2} \epsilon_0 E^2\).
\(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 90 \times 10^{-4}}{0.0025} = 3.186 \times 10^{-11} \, \text{F}\).
Energy: \(U = \frac{1}{2} C V^2 = \frac{1}{2} \times 3.186 \times 10^{-11} \times (400)^2 = 2.55 \times 10^{-6} \, \text{J}\).
(b) Energy density:
Volume = \(A d = 90 \times 10^{-4} \times 0.0025 = 2.25 \times 10^{-5} \, \text{m}^3\).
\(u = \frac{U}{\text{vol}} = \frac{2.55 \times 10^{-6}}{2.25 \times 10^{-5}} = 0.113 \, \text{J/m}^3\).
Also \(u = \frac{1}{2} \epsilon_0 E^2\), where \(E = V/d = 400/0.0025 = 1.6 \times 10^5 \, \text{V/m}\).
Answer: (a) 2.55 µJ, (b) u = 0.113 J/m³, \(u = \frac{1}{2} \epsilon_0 E^2\).
Question 2.27
A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Answer & Explanation:
Step 1: Initial energy
\(U_i = \frac{1}{2} \times 4 \times 10^{-6} \times (200)^2 = 8 \times 10^{-2} \, \text{J}\).
Step 2: After connection
Total C = 6 µF, charge Q = 4×10⁻⁶ × 200 = 8×10⁻⁴ C.
Common V = Q / C_total = 8×10⁻⁴ / 6×10⁻⁶ = 133.33 V.
Final energy: \(U_f = \frac{1}{2} \times 6 \times 10^{-6} \times (133.33)^2 ≈ 5.333 \times 10^{-2} \, \text{J}\).
Step 3: Energy lost
\(\Delta U = U_i - U_f = 0.08 - 0.05333 = 0.02667 \, \text{J}\).
Answer: 26.67 mJ.
\(U_i = \frac{1}{2} \times 4 \times 10^{-6} \times (200)^2 = 8 \times 10^{-2} \, \text{J}\).
Step 2: After connection
Total C = 6 µF, charge Q = 4×10⁻⁶ × 200 = 8×10⁻⁴ C.
Common V = Q / C_total = 8×10⁻⁴ / 6×10⁻⁶ = 133.33 V.
Final energy: \(U_f = \frac{1}{2} \times 6 \times 10^{-6} \times (133.33)^2 ≈ 5.333 \times 10^{-2} \, \text{J}\).
Step 3: Energy lost
\(\Delta U = U_i - U_f = 0.08 - 0.05333 = 0.02667 \, \text{J}\).
Answer: 26.67 mJ.
Question 2.28
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) \(QE\), where \(Q\) is the charge on the capacitor, and \(E\) is the magnitude of electric field between the plates. Explain the origin of the factor \(½\).
Answer & Explanation:
Force on plate due to field of other plate:
Field at one plate due to the other = \(\frac{\sigma}{2\epsilon_0} = \frac{E}{2}\).
Force = charge × field = \(Q \times \frac{E}{2} = \frac{1}{2} QE\).
Factor ½ appears because a plate does not exert a net force on itself; the field acting on it is only due to the opposite plate.
Answer: Derived.
Field at one plate due to the other = \(\frac{\sigma}{2\epsilon_0} = \frac{E}{2}\).
Force = charge × field = \(Q \times \frac{E}{2} = \frac{1}{2} QE\).
Factor ½ appears because a plate does not exert a net force on itself; the field acting on it is only due to the opposite plate.
Answer: Derived.
Question 2.29
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show that the capacitance of a spherical capacitor is given by \(C = \frac{4\pi\epsilon_0 r_1 r_2}{r_1 - r_2}\) where \(r_1\) and \(r_2\) are the radii of outer and inner spheres, respectively.
Answer & Explanation:
Derivation:
Potential difference between spheres: \(V = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{r_2} - \frac{1}{r_1} \right)\)
Capacitance \(C = Q/V = \frac{4\pi\epsilon_0 r_1 r_2}{r_1 - r_2}\).
Answer: Derived.
Potential difference between spheres: \(V = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{r_2} - \frac{1}{r_1} \right)\)
Capacitance \(C = Q/V = \frac{4\pi\epsilon_0 r_1 r_2}{r_1 - r_2}\).
Answer: Derived.
Question 2.30
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Answer & Explanation:
(a) Capacitance with dielectric:
\(C = \frac{4\pi\epsilon_0 K r_1 r_2}{r_2 - r_1}\)
= \( \frac{4\pi \times 8.85 \times 10^{-12} \times 32 \times 0.12 \times 0.13}{0.01} \)
≈ 5.56 × 10⁻⁹ F = 5.56 nF.
(b) Potential:
\(V = Q/C = 2.5 \times 10^{-6} / 5.56 \times 10^{-9} ≈ 450 \, \text{V}\).
(c) Isolated sphere: C = 4πϵ₀ r ≈ 13.3 pF, much smaller.
Answer: (a) ~5.56 nF, (b) ~450 V, (c) much smaller.
\(C = \frac{4\pi\epsilon_0 K r_1 r_2}{r_2 - r_1}\)
= \( \frac{4\pi \times 8.85 \times 10^{-12} \times 32 \times 0.12 \times 0.13}{0.01} \)
≈ 5.56 × 10⁻⁹ F = 5.56 nF.
(b) Potential:
\(V = Q/C = 2.5 \times 10^{-6} / 5.56 \times 10^{-9} ≈ 450 \, \text{V}\).
(c) Isolated sphere: C = 4πϵ₀ r ≈ 13.3 pF, much smaller.
Answer: (a) ~5.56 nF, (b) ~450 V, (c) much smaller.
Question 2.31
Answer carefully:
(a) Two large conducting spheres carrying charges \(Q_1\) and \(Q_2\) are brought close to each other. Is the magnitude of electrostatic force between them exactly given by \(Q_1 Q_2 / 4\pi\epsilon_0 r^2\), where \(r\) is the distance between their centres?
(b) If Coulomb’s law involved \(1/r^3\) dependence (instead of \(1/r^2\)), would Gauss’s law be still true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
(a) Two large conducting spheres carrying charges \(Q_1\) and \(Q_2\) are brought close to each other. Is the magnitude of electrostatic force between them exactly given by \(Q_1 Q_2 / 4\pi\epsilon_0 r^2\), where \(r\) is the distance between their centres?
(b) If Coulomb’s law involved \(1/r^3\) dependence (instead of \(1/r^2\)), would Gauss’s law be still true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Answer & Explanation:
(a) Not exactly, because charges redistribute.
(b) No, Gauss’s law relies on 1/r².
(c) Yes, if initial velocity is zero.
(d) Zero for any closed path (conservative).
(e) No, potential continuous, field discontinuous.
(f) Capacitance relative to infinity: C = Q/V.
(g) Water molecules have high polarizability.
Answer: As above.
(b) No, Gauss’s law relies on 1/r².
(c) Yes, if initial velocity is zero.
(d) Zero for any closed path (conservative).
(e) No, potential continuous, field discontinuous.
(f) Capacitance relative to infinity: C = Q/V.
(g) Water molecules have high polarizability.
Answer: As above.
Question 2.32
A cylindrical capacitor has two co‑axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
Answer & Explanation:
Capacitance per length:
\(C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}\)
\(= \frac{2\pi \times 8.85 \times 10^{-12} \times 0.15}{\ln(1.5/1.4)}\)
≈ 1.16 × 10⁻¹⁰ F = 116 pF.
Potential:
\(V = Q/C = 3.5 \times 10^{-6} / 1.16 \times 10^{-10} ≈ 30.2 \, \text{kV}\).
Answer: C ≈ 116 pF, V ≈ 30.2 kV.
\(C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}\)
\(= \frac{2\pi \times 8.85 \times 10^{-12} \times 0.15}{\ln(1.5/1.4)}\)
≈ 1.16 × 10⁻¹⁰ F = 116 pF.
Potential:
\(V = Q/C = 3.5 \times 10^{-6} / 1.16 \times 10^{-10} ≈ 30.2 \, \text{kV}\).
Answer: C ≈ 116 pF, V ≈ 30.2 kV.
Question 2.33
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ V/m. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Answer & Explanation:
Step 1: Max allowed field
\(E_{\max} = 0.1 \times 10^7 = 10^6 \, \text{V/m}\).
Step 2: Plate separation
\(d \geq V / E_{\max} = 1000 / 10^6 = 10^{-3} \, \text{m} = 1 \, \text{mm}\).
Step 3: Area
\(C = \frac{K \epsilon_0 A}{d}\) → \(A = \frac{C d}{K \epsilon_0} = \frac{50 \times 10^{-12} \times 0.001}{3 \times 8.85 \times 10^{-12}} ≈ 1.88 \times 10^{-3} \, \text{m}^2\).
Answer: ~1.88 × 10⁻³ m² = 18.8 cm².
\(E_{\max} = 0.1 \times 10^7 = 10^6 \, \text{V/m}\).
Step 2: Plate separation
\(d \geq V / E_{\max} = 1000 / 10^6 = 10^{-3} \, \text{m} = 1 \, \text{mm}\).
Step 3: Area
\(C = \frac{K \epsilon_0 A}{d}\) → \(A = \frac{C d}{K \epsilon_0} = \frac{50 \times 10^{-12} \times 0.001}{3 \times 8.85 \times 10^{-12}} ≈ 1.88 \times 10^{-3} \, \text{m}^2\).
Answer: ~1.88 × 10⁻³ m² = 18.8 cm².
Question 2.34
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z‑direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
(a) a constant electric field in the z‑direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer & Explanation:
(a) Planes perpendicular to z‑axis (parallel to xy‑plane).
(b) Non‑equispaced planes perpendicular to z.
(c) Concentric spheres centered at origin.
(d) Cylindrical surfaces around wires, or periodic pattern in plane.
Answer: As above.
(b) Non‑equispaced planes perpendicular to z.
(c) Concentric spheres centered at origin.
(d) Cylindrical surfaces around wires, or periodic pattern in plane.
Answer: As above.
Question 2.35
In a Van de Graaff type generator a spherical metal shell is to be a \(15 \times 10^6 \, \text{V}\) electrode. The dielectric strength of the gas surrounding the electrode is \(5 \times 10^7 \, \text{V/m}\). What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Answer & Explanation:
Field at surface: \(E = \frac{k Q}{R^2} = \frac{V}{R}\) (since V = kQ/R).
Set \(E \leq 5 \times 10^7\).
\(V/R \leq 5 \times 10^7\) → \(R \geq V / (5 \times 10^7) = \frac{15 \times 10^6}{5 \times 10^7} = 0.3 \, \text{m}\).
Answer: Minimum radius 0.3 m = 30 cm.
Set \(E \leq 5 \times 10^7\).
\(V/R \leq 5 \times 10^7\) → \(R \geq V / (5 \times 10^7) = \frac{15 \times 10^6}{5 \times 10^7} = 0.3 \, \text{m}\).
Answer: Minimum radius 0.3 m = 30 cm.
Question 2.36
A small sphere of radius \(r_1\) and charge \(q_1\) is enclosed by a spherical shell of radius \(r_2\) and charge \(q_2\). Show that if \(q_1\) is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge \(q_2\) on the shell is.
Answer & Explanation:
Potential of inner sphere: \(V_1 = \frac{k q_1}{r_1} + \frac{k q_2}{r_2}\).
Potential of shell: \(V_2 = \frac{k q_1}{r_2} + \frac{k q_2}{r_2}\).
Difference: \(V_1 - V_2 = k q_1 \left( \frac{1}{r_1} - \frac{1}{r_2} \right) > 0\) if q₁>0.
So V₁ > V₂ → charge flows from sphere to shell.
Answer: Proved.
Potential of shell: \(V_2 = \frac{k q_1}{r_2} + \frac{k q_2}{r_2}\).
Difference: \(V_1 - V_2 = k q_1 \left( \frac{1}{r_1} - \frac{1}{r_2} \right) > 0\) if q₁>0.
So V₁ > V₂ → charge flows from sphere to shell.
Answer: Proved.
Question 2.37
Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V/m. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m². Will he get an electric shock if he touches the metal sheet next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V/m. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m². Will he get an electric shock if he touches the metal sheet next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
Answer & Explanation:
(a) Potential difference is high, but current small due to air resistance; body not good conductor in open.
(b) Possibly yes if sheet charged by atmospheric field.
(c) Continuous charging by thunderstorms balances leakage.
(d) Light, heat, sound, radio waves, ionization.
Answer: As above.
(b) Possibly yes if sheet charged by atmospheric field.
(c) Continuous charging by thunderstorms balances leakage.
(d) Light, heat, sound, radio waves, ionization.
Answer: As above.
📘 Exam Preparation Tip:
These exercise questions will help you understand energy concepts in electrostatics and capacitor applications. You'll learn to calculate electrostatic potential and potential energy, determine capacitance for various geometries, analyze capacitor combinations (series/parallel), calculate energy stored in capacitors, and understand dielectric effects on capacitance. Crucial for circuit analysis and understanding energy storage devices in real-world applications.