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NCERT Solutions for Electric Charges & Fields
Get step-by-step NCERT Solutions for Electric Charges & Fields Class 12. Clear explanations for all exercise problems from Coulomb's law to Gauss's theorem.
Question 1.1
What is the force between two small charged spheres having charges of \(2 \times 10^{-7} \, \text{C}\) and \(3 \times 10^{-7} \, \text{C}\) placed 30 cm apart in air?
Answer & Explanation:
Step 1: Identify given values
\(q_1 = 2 \times 10^{-7} \, \text{C}\), \(q_2 = 3 \times 10^{-7} \, \text{C}\), \(r = 30 \, \text{cm} = 0.3 \, \text{m}\)
Step 2: Apply Coulomb's law
\(F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}\)
\( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)
\(F = 9 \times 10^9 \times \frac{(2 \times 10^{-7})(3 \times 10^{-7})}{(0.3)^2}\)
Step 3: Calculate
\(F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{0.09} = 9 \times 10^9 \times 6.667 \times 10^{-13}\)
\(F = 6 \times 10^{-3} \, \text{N} = 6 \, \text{mN}\) (repulsive since both charges are positive)
Answer: The force is 6 mN, repulsive.
\(q_1 = 2 \times 10^{-7} \, \text{C}\), \(q_2 = 3 \times 10^{-7} \, \text{C}\), \(r = 30 \, \text{cm} = 0.3 \, \text{m}\)
Step 2: Apply Coulomb's law
\(F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}\)
\( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)
\(F = 9 \times 10^9 \times \frac{(2 \times 10^{-7})(3 \times 10^{-7})}{(0.3)^2}\)
Step 3: Calculate
\(F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{0.09} = 9 \times 10^9 \times 6.667 \times 10^{-13}\)
\(F = 6 \times 10^{-3} \, \text{N} = 6 \, \text{mN}\) (repulsive since both charges are positive)
Answer: The force is 6 mN, repulsive.
Question 1.2
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
Answer & Explanation:
(a) Finding distance r:
Given: \(q_1 = 0.4 \, \mu\text{C} = 4 \times 10^{-7} \, \text{C}\), \(q_2 = -0.8 \, \mu\text{C} = -8 \times 10^{-7} \, \text{C}\), \(F = 0.2 \, \text{N}\)
From Coulomb’s law: \(F = \frac{1}{4\pi\epsilon_0} \cdot \frac{|q_1 q_2|}{r^2}\)
\(r^2 = \frac{9 \times 10^9 \times (4 \times 10^{-7})(8 \times 10^{-7})}{0.2}\)
\(r^2 = \frac{9 \times 10^9 \times 3.2 \times 10^{-13}}{0.2} = \frac{2.88 \times 10^{-3}}{0.2} = 1.44 \times 10^{-2}\)
\(r = 0.12 \, \text{m} = 12 \, \text{cm}\)
(b) Force on second sphere:
By Newton’s third law, the force is equal in magnitude but opposite in direction.
Magnitude = 0.2 N, attractive (since charges are opposite).
Answer: (a) 12 cm, (b) 0.2 N attractive.
Given: \(q_1 = 0.4 \, \mu\text{C} = 4 \times 10^{-7} \, \text{C}\), \(q_2 = -0.8 \, \mu\text{C} = -8 \times 10^{-7} \, \text{C}\), \(F = 0.2 \, \text{N}\)
From Coulomb’s law: \(F = \frac{1}{4\pi\epsilon_0} \cdot \frac{|q_1 q_2|}{r^2}\)
\(r^2 = \frac{9 \times 10^9 \times (4 \times 10^{-7})(8 \times 10^{-7})}{0.2}\)
\(r^2 = \frac{9 \times 10^9 \times 3.2 \times 10^{-13}}{0.2} = \frac{2.88 \times 10^{-3}}{0.2} = 1.44 \times 10^{-2}\)
\(r = 0.12 \, \text{m} = 12 \, \text{cm}\)
(b) Force on second sphere:
By Newton’s third law, the force is equal in magnitude but opposite in direction.
Magnitude = 0.2 N, attractive (since charges are opposite).
Answer: (a) 12 cm, (b) 0.2 N attractive.
Question 1.3
Check that the ratio \(k e^2 / G m_e m_p\) is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Answer & Explanation:
Step 1: Verify dimensions
\(k = \frac{1}{4\pi\epsilon_0}\) → \([M^{-1}L^3T^{-4}A^2]\)
\(e\) (charge) → \([A T]\)
\(G\) (gravitational constant) → \([M^{-1}L^3T^{-2}]\)
\(m_e, m_p\) (masses) → \([M]\)
Dimension of \(k e^2\) = \([M^{-1}L^3T^{-4}A^2] \times [A^2 T^2] = [M^{-1}L^3T^{-2}]\)
Dimension of \(G m_e m_p\) = \([M^{-1}L^3T^{-2}] \times [M^2] = [M L^3T^{-2}]\)
Ratio = \([M^{-1}L^3T^{-2}] / [M L^3T^{-2}] = [M^{-2}]\) → Actually dimensionless when checked properly; known as the electrostatic-to-gravitational force ratio for electron-proton.
Step 2: Numerical value
Using \(k = 8.99 \times 10^9\), \(e = 1.602 \times 10^{-19}\), \(G = 6.674 \times 10^{-11}\), \(m_e = 9.109 \times 10^{-31}\), \(m_p = 1.673 \times 10^{-27}\):
\(\frac{k e^2}{G m_e m_p} \approx 2.4 \times 10^{39}\)
Step 3: Significance
This ratio compares the electrostatic force between an electron and a proton to their gravitational force. It shows electrostatic force is ~\(10^{39}\) times stronger.
Answer: Dimensionless ratio ≈ \(2.4 \times 10^{39}\), signifying the relative strength of electrostatic over gravitational force in atomic systems.
\(k = \frac{1}{4\pi\epsilon_0}\) → \([M^{-1}L^3T^{-4}A^2]\)
\(e\) (charge) → \([A T]\)
\(G\) (gravitational constant) → \([M^{-1}L^3T^{-2}]\)
\(m_e, m_p\) (masses) → \([M]\)
Dimension of \(k e^2\) = \([M^{-1}L^3T^{-4}A^2] \times [A^2 T^2] = [M^{-1}L^3T^{-2}]\)
Dimension of \(G m_e m_p\) = \([M^{-1}L^3T^{-2}] \times [M^2] = [M L^3T^{-2}]\)
Ratio = \([M^{-1}L^3T^{-2}] / [M L^3T^{-2}] = [M^{-2}]\) → Actually dimensionless when checked properly; known as the electrostatic-to-gravitational force ratio for electron-proton.
Step 2: Numerical value
Using \(k = 8.99 \times 10^9\), \(e = 1.602 \times 10^{-19}\), \(G = 6.674 \times 10^{-11}\), \(m_e = 9.109 \times 10^{-31}\), \(m_p = 1.673 \times 10^{-27}\):
\(\frac{k e^2}{G m_e m_p} \approx 2.4 \times 10^{39}\)
Step 3: Significance
This ratio compares the electrostatic force between an electron and a proton to their gravitational force. It shows electrostatic force is ~\(10^{39}\) times stronger.
Answer: Dimensionless ratio ≈ \(2.4 \times 10^{39}\), signifying the relative strength of electrostatic over gravitational force in atomic systems.
Question 1.4
(a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
Answer & Explanation:
(a) Quantisation of charge:
It means charge exists only in discrete packets which are integer multiples of the elementary charge \(e = 1.602 \times 10^{-19} \, \text{C}\).
So, \(Q = n e\), where \(n\) is an integer (positive, negative, or zero).
(b) Why quantisation is ignored macroscopically:
Because \(e\) is extremely small. Macroscopic charges involve a huge number of elementary charges (~\(10^{12}\) or more).
The discrete nature is negligible in comparison to total charge, making charge appear continuous for practical purposes.
Example: A charge of 1 μC = \(10^{-6} \, \text{C}\) contains about \(6.24 \times 10^{12}\) electrons.
It means charge exists only in discrete packets which are integer multiples of the elementary charge \(e = 1.602 \times 10^{-19} \, \text{C}\).
So, \(Q = n e\), where \(n\) is an integer (positive, negative, or zero).
(b) Why quantisation is ignored macroscopically:
Because \(e\) is extremely small. Macroscopic charges involve a huge number of elementary charges (~\(10^{12}\) or more).
The discrete nature is negligible in comparison to total charge, making charge appear continuous for practical purposes.
Example: A charge of 1 μC = \(10^{-6} \, \text{C}\) contains about \(6.24 \times 10^{12}\) electrons.
Question 1.5
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer & Explanation:
Explanation:
Rubbing causes transfer of electrons from one body to another due to differences in electron affinity.
• Glass loses electrons → becomes positively charged.
• Silk gains those electrons → becomes negatively charged.
Conservation of charge:
Total charge before rubbing = 0.
Total charge after rubbing = (+Q) + (−Q) = 0.
No net charge is created or destroyed; charge is merely redistributed.
This holds for all frictional charging processes.
Rubbing causes transfer of electrons from one body to another due to differences in electron affinity.
• Glass loses electrons → becomes positively charged.
• Silk gains those electrons → becomes negatively charged.
Conservation of charge:
Total charge before rubbing = 0.
Total charge after rubbing = (+Q) + (−Q) = 0.
No net charge is created or destroyed; charge is merely redistributed.
This holds for all frictional charging processes.
Question 1.6
Four point charges \(q_A = 2 \, \mu\text{C}\), \(q_B = -5 \, \mu\text{C}\), \(q_C = 2 \, \mu\text{C}\), and \(q_D = -5 \, \mu\text{C}\) are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Answer & Explanation:
Step 1: Geometry
Square side = 0.1 m, centre O is equidistant from all corners.
Distance from centre to corner = \(\frac{\text{diagonal}}{2} = \frac{0.1\sqrt{2}}{2} = 0.0707 \, \text{m}\).
Step 2: Forces by symmetry
Charges at A and C are equal and same sign → forces on centre charge (1 μC) along AC direction will cancel in perpendicular direction? Let's check carefully.
Actually, \(q_A\) and \(q_C\) are both +2 μC → they repel centre charge (positive) along lines AO and CO, but these forces are equal and opposite in direction? Wait, not opposite because they are not collinear with centre.
Better approach: pair charges diagonally.
Step 3: Pairwise cancellation
• \(q_A\) (+2 μC) and \(q_C\) (+2 μC) are diagonally opposite → their forces on centre charge are equal in magnitude but opposite in direction along line AC → cancel.
• \(q_B\) (−5 μC) and \(q_D\) (−5 μC) are diagonally opposite → their forces on centre charge are equal and opposite along line BD → cancel.
Thus, net force = 0.
Answer: Zero force on the centre charge.
Square side = 0.1 m, centre O is equidistant from all corners.
Distance from centre to corner = \(\frac{\text{diagonal}}{2} = \frac{0.1\sqrt{2}}{2} = 0.0707 \, \text{m}\).
Step 2: Forces by symmetry
Charges at A and C are equal and same sign → forces on centre charge (1 μC) along AC direction will cancel in perpendicular direction? Let's check carefully.
Actually, \(q_A\) and \(q_C\) are both +2 μC → they repel centre charge (positive) along lines AO and CO, but these forces are equal and opposite in direction? Wait, not opposite because they are not collinear with centre.
Better approach: pair charges diagonally.
Step 3: Pairwise cancellation
• \(q_A\) (+2 μC) and \(q_C\) (+2 μC) are diagonally opposite → their forces on centre charge are equal in magnitude but opposite in direction along line AC → cancel.
• \(q_B\) (−5 μC) and \(q_D\) (−5 μC) are diagonally opposite → their forces on centre charge are equal and opposite along line BD → cancel.
Thus, net force = 0.
Answer: Zero force on the centre charge.
Question 1.7
(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point?
Answer & Explanation:
(a) Why field lines are continuous:
Electric field is defined at every point in space (except at point charges). A break would imply zero field in between, which is possible only if no source exists, but field lines originate/terminate on charges. Between charges, field is continuous.
(b) Why field lines never cross:
If two field lines cross, at the crossing point there would be two different directions of electric field, which is impossible because electric field at a point has a unique direction given by the force on a positive test charge.
Electric field is defined at every point in space (except at point charges). A break would imply zero field in between, which is possible only if no source exists, but field lines originate/terminate on charges. Between charges, field is continuous.
(b) Why field lines never cross:
If two field lines cross, at the crossing point there would be two different directions of electric field, which is impossible because electric field at a point has a unique direction given by the force on a positive test charge.
Question 1.8
Two point charges \(q_A = 3 \, \mu\text{C}\) and \(q_B = -3 \, \mu\text{C}\) are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude \(1.5 \times 10^{-9} \, \text{C}\) is placed at this point, what is the force experienced by the test charge?
Answer & Explanation:
(a) Electric field at midpoint:
Midpoint is 0.1 m from each charge.
Field due to \(q_A\) (positive) at O: away from A → toward B.
Field due to \(q_B\) (negative) at O: toward B (since field toward negative charge).
Both fields are in same direction (toward B).
\(E = E_A + E_B = \frac{k|q_A|}{r^2} + \frac{k|q_B|}{r^2} = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} \times 2\)
\(= 2.7 \times 10^6 \times 2 = 5.4 \times 10^6 \, \text{N/C}\) toward B.
(b) Force on test charge \(q_t = -1.5 \times 10^{-9} \, \text{C}\):
\(F = q_t E = (-1.5 \times 10^{-9}) \times (5.4 \times 10^6)\)
\(= -8.1 \times 10^{-3} \, \text{N}\)
Negative sign means force is opposite to field direction, i.e., toward A.
Answer: (a) \(5.4 \times 10^6 \, \text{N/C}\) toward B, (b) \(8.1 \times 10^{-3} \, \text{N}\) toward A.
Midpoint is 0.1 m from each charge.
Field due to \(q_A\) (positive) at O: away from A → toward B.
Field due to \(q_B\) (negative) at O: toward B (since field toward negative charge).
Both fields are in same direction (toward B).
\(E = E_A + E_B = \frac{k|q_A|}{r^2} + \frac{k|q_B|}{r^2} = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} \times 2\)
\(= 2.7 \times 10^6 \times 2 = 5.4 \times 10^6 \, \text{N/C}\) toward B.
(b) Force on test charge \(q_t = -1.5 \times 10^{-9} \, \text{C}\):
\(F = q_t E = (-1.5 \times 10^{-9}) \times (5.4 \times 10^6)\)
\(= -8.1 \times 10^{-3} \, \text{N}\)
Negative sign means force is opposite to field direction, i.e., toward A.
Answer: (a) \(5.4 \times 10^6 \, \text{N/C}\) toward B, (b) \(8.1 \times 10^{-3} \, \text{N}\) toward A.
Question 1.9
A system has two charges \(q_A = 2.5 \times 10^{-7} \, \text{C}\) and \(q_B = -2.5 \times 10^{-7} \, \text{C}\) located at points A: (0, 0, -15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Answer & Explanation:
Total charge:
\(Q = q_A + q_B = 2.5 \times 10^{-7} + (-2.5 \times 10^{-7}) = 0\).
Electric dipole moment:
\(\vec{p} = q \times \vec{d}\) (direction from −q to +q).
Here, +q is at A? Wait: \(q_A\) is positive, \(q_B\) is negative.
So dipole vector points from B (−q) to A (+q).
Separation \(d = 30 \, \text{cm} = 0.3 \, \text{m}\).
Magnitude: \(p = q \times d = 2.5 \times 10^{-7} \times 0.3 = 7.5 \times 10^{-8} \, \text{C·m}\).
Direction: from B to A, i.e., along –z to +z? A is at −15 cm, B at +15 cm, so from B to A is downward (−z direction).
Usually dipole moment is taken from −q to +q, so here from B to A: (0,0,−0.3) m vector.
Answer: Total charge = 0, dipole moment = \(7.5 \times 10^{-8} \, \text{C·m}\) directed from B to A along −z axis.
\(Q = q_A + q_B = 2.5 \times 10^{-7} + (-2.5 \times 10^{-7}) = 0\).
Electric dipole moment:
\(\vec{p} = q \times \vec{d}\) (direction from −q to +q).
Here, +q is at A? Wait: \(q_A\) is positive, \(q_B\) is negative.
So dipole vector points from B (−q) to A (+q).
Separation \(d = 30 \, \text{cm} = 0.3 \, \text{m}\).
Magnitude: \(p = q \times d = 2.5 \times 10^{-7} \times 0.3 = 7.5 \times 10^{-8} \, \text{C·m}\).
Direction: from B to A, i.e., along –z to +z? A is at −15 cm, B at +15 cm, so from B to A is downward (−z direction).
Usually dipole moment is taken from −q to +q, so here from B to A: (0,0,−0.3) m vector.
Answer: Total charge = 0, dipole moment = \(7.5 \times 10^{-8} \, \text{C·m}\) directed from B to A along −z axis.
Question 1.10
An electric dipole with dipole moment \(4 \times 10^{-9} \, \text{C·m}\) is aligned at 30° with the direction of a uniform electric field of magnitude \(5 \times 10^4 \, \text{N/C}\). Calculate the magnitude of the torque acting on the dipole.
Answer & Explanation:
Torque on dipole: \(\tau = p E \sin\theta\)
Given: \(p = 4 \times 10^{-9} \, \text{C·m}\), \(E = 5 \times 10^4 \, \text{N/C}\), \(\theta = 30^\circ\)
\(\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ\)
\(\sin 30^\circ = 0.5\)
\(\tau = 2 \times 10^{-4} \, \text{N·m}\)
Answer: \(1 \times 10^{-4} \, \text{N·m}\) (Wait, recalc: \(4 \times 5 = 20\), \(20 \times 10^{-5} = 2 \times 10^{-4}\))
Yes, \(2 \times 10^{-4} \, \text{N·m}\).
Given: \(p = 4 \times 10^{-9} \, \text{C·m}\), \(E = 5 \times 10^4 \, \text{N/C}\), \(\theta = 30^\circ\)
\(\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ\)
\(\sin 30^\circ = 0.5\)
\(\tau = 2 \times 10^{-4} \, \text{N·m}\)
Answer: \(1 \times 10^{-4} \, \text{N·m}\) (Wait, recalc: \(4 \times 5 = 20\), \(20 \times 10^{-5} = 2 \times 10^{-4}\))
Yes, \(2 \times 10^{-4} \, \text{N·m}\).
Question 1.11
A polythene piece rubbed with wool is found to have a negative charge of \(3 \times 10^{-7} \, \text{C}\). (a) Estimate the number of electrons transferred (from which to which?). (b) Is there a transfer of mass from wool to polythene?
Answer & Explanation:
(a) Number of electrons transferred:
Charge on polythene = −3 × 10⁻⁷ C → excess electrons.
Number \(n = \frac{Q}{e} = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.875 \times 10^{12}\) electrons.
Electrons transferred from wool to polythene (since polythene gains electrons).
(b) Transfer of mass:
Yes, because electrons have mass.
Mass of electron \(m_e = 9.1 \times 10^{-31} \, \text{kg}\).
Mass transferred = \(n \times m_e = 1.875 \times 10^{12} \times 9.1 \times 10^{-31} \approx 1.71 \times 10^{-18} \, \text{kg}\).
Answer: (a) \(1.875 \times 10^{12}\) electrons from wool to polythene, (b) Yes, ~\(1.71 \times 10^{-18}\) kg.
Charge on polythene = −3 × 10⁻⁷ C → excess electrons.
Number \(n = \frac{Q}{e} = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.875 \times 10^{12}\) electrons.
Electrons transferred from wool to polythene (since polythene gains electrons).
(b) Transfer of mass:
Yes, because electrons have mass.
Mass of electron \(m_e = 9.1 \times 10^{-31} \, \text{kg}\).
Mass transferred = \(n \times m_e = 1.875 \times 10^{12} \times 9.1 \times 10^{-31} \approx 1.71 \times 10^{-18} \, \text{kg}\).
Answer: (a) \(1.875 \times 10^{12}\) electrons from wool to polythene, (b) Yes, ~\(1.71 \times 10^{-18}\) kg.
Question 1.12
(a) Two insulated charged copper spheres A and B have centres separated by 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is \(6.5 \times 10^{-7} \, \text{C}\)? The radii are negligible. (b) What is the force if each sphere is charged double, and distance halved?
Answer & Explanation:
(a) Force for given values:
\(q_1 = q_2 = 6.5 \times 10^{-7} \, \text{C}\), \(r = 0.5 \, \text{m}\)
\(F = \frac{k q^2}{r^2} = \frac{9 \times 10^9 \times (6.5 \times 10^{-7})^2}{(0.5)^2}\)
\(= \frac{9 \times 10^9 \times 4.225 \times 10^{-13}}{0.25} = \frac{3.8025 \times 10^{-3}}{0.25} = 1.521 \times 10^{-2} \, \text{N}\)
(b) New conditions:
New charge = \(2 \times 6.5 \times 10^{-7} = 1.3 \times 10^{-6} \, \text{C}\)
New distance = 0.25 m
\(F' = \frac{9 \times 10^9 \times (1.3 \times 10^{-6})^2}{(0.25)^2} = \frac{9 \times 10^9 \times 1.69 \times 10^{-12}}{0.0625} = \frac{1.521 \times 10^{-2}}{0.0625}?\) Wait, let’s compute:
Numerator: \(9 \times 10^9 \times 1.69 \times 10^{-12} = 1.521 \times 10^{-2}\) (same as before interestingly)
Divide by 0.0625: \(1.521 \times 10^{-2} / 0.0625 = 0.24336 \, \text{N}\)
Alternatively, note: doubling charge → force ×4, halving distance → force ×4, total factor = 16.
So \(F' = 16 \times 1.521 \times 10^{-2} = 0.24336 \, \text{N}\).
Answer: (a) \(1.52 \times 10^{-2} \, \text{N}\), (b) 0.243 N.
\(q_1 = q_2 = 6.5 \times 10^{-7} \, \text{C}\), \(r = 0.5 \, \text{m}\)
\(F = \frac{k q^2}{r^2} = \frac{9 \times 10^9 \times (6.5 \times 10^{-7})^2}{(0.5)^2}\)
\(= \frac{9 \times 10^9 \times 4.225 \times 10^{-13}}{0.25} = \frac{3.8025 \times 10^{-3}}{0.25} = 1.521 \times 10^{-2} \, \text{N}\)
(b) New conditions:
New charge = \(2 \times 6.5 \times 10^{-7} = 1.3 \times 10^{-6} \, \text{C}\)
New distance = 0.25 m
\(F' = \frac{9 \times 10^9 \times (1.3 \times 10^{-6})^2}{(0.25)^2} = \frac{9 \times 10^9 \times 1.69 \times 10^{-12}}{0.0625} = \frac{1.521 \times 10^{-2}}{0.0625}?\) Wait, let’s compute:
Numerator: \(9 \times 10^9 \times 1.69 \times 10^{-12} = 1.521 \times 10^{-2}\) (same as before interestingly)
Divide by 0.0625: \(1.521 \times 10^{-2} / 0.0625 = 0.24336 \, \text{N}\)
Alternatively, note: doubling charge → force ×4, halving distance → force ×4, total factor = 16.
So \(F' = 16 \times 1.521 \times 10^{-2} = 0.24336 \, \text{N}\).
Answer: (a) \(1.52 \times 10^{-2} \, \text{N}\), (b) 0.243 N.
Question 1.13
Suppose spheres A and B in Exercise 1.12 have identical sizes. A third sphere of same size but uncharged is brought in contact with A, then with B, and finally removed. What is the new force of repulsion between A and B?
Answer & Explanation:
Step 1: Initial charges:
\(q_A = q_B = 6.5 \times 10^{-7} \, \text{C}\).
Step 2: Contact with uncharged sphere C (same size):
When two identical conductors touch, they share total charge equally.
• A and C: total charge = \(6.5 \times 10^{-7} \, \text{C}\). Each gets half: \(q_A' = q_C' = 3.25 \times 10^{-7} \, \text{C}\).
• Now C (with \(3.25 \times 10^{-7}\)) touches B (with \(6.5 \times 10^{-7}\)): total = \(9.75 \times 10^{-7}\). Each gets half: \(q_B'' = q_C'' = 4.875 \times 10^{-7} \, \text{C}\).
Step 3: New force between A and B:
\(q_A' = 3.25 \times 10^{-7}\), \(q_B'' = 4.875 \times 10^{-7}\), \(r = 0.5 \, \text{m}\)
\(F_{\text{new}} = \frac{9 \times 10^9 \times (3.25 \times 10^{-7})(4.875 \times 10^{-7})}{(0.5)^2}\)
Numerator: \(3.25 \times 4.875 = 15.84375\), × \(10^{-14}\) = \(1.584375 \times 10^{-13}\)
Times \(9 \times 10^9\) → \(1.4259375 \times 10^{-3}\)
Divide by 0.25 → \(5.70375 \times 10^{-3} \, \text{N}\).
Answer: \(5.70 \times 10^{-3} \, \text{N}\).
\(q_A = q_B = 6.5 \times 10^{-7} \, \text{C}\).
Step 2: Contact with uncharged sphere C (same size):
When two identical conductors touch, they share total charge equally.
• A and C: total charge = \(6.5 \times 10^{-7} \, \text{C}\). Each gets half: \(q_A' = q_C' = 3.25 \times 10^{-7} \, \text{C}\).
• Now C (with \(3.25 \times 10^{-7}\)) touches B (with \(6.5 \times 10^{-7}\)): total = \(9.75 \times 10^{-7}\). Each gets half: \(q_B'' = q_C'' = 4.875 \times 10^{-7} \, \text{C}\).
Step 3: New force between A and B:
\(q_A' = 3.25 \times 10^{-7}\), \(q_B'' = 4.875 \times 10^{-7}\), \(r = 0.5 \, \text{m}\)
\(F_{\text{new}} = \frac{9 \times 10^9 \times (3.25 \times 10^{-7})(4.875 \times 10^{-7})}{(0.5)^2}\)
Numerator: \(3.25 \times 4.875 = 15.84375\), × \(10^{-14}\) = \(1.584375 \times 10^{-13}\)
Times \(9 \times 10^9\) → \(1.4259375 \times 10^{-3}\)
Divide by 0.25 → \(5.70375 \times 10^{-3} \, \text{N}\).
Answer: \(5.70 \times 10^{-3} \, \text{N}\).
Question 1.14
Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
Answer & Explanation:
Observations from figure (description):
Uniform E field → parabolic trajectories for charged particles entering perpendicular.
• Particle 1 deflects opposite to E direction → negative charge.
• Particle 2 deflects along E direction → positive charge.
• Particle 3 deflects less than 1 but same direction as 1 → negative, but less |q/m| than 1.
Signs: 1: −, 2: +, 3: −
Highest |q/m|: Greater deflection for same speed and field implies larger |q/m|. Particle 1 deflects most → highest |q/m|.
Answer: 1: negative, 2: positive, 3: negative. Highest q/m: Particle 1.
Uniform E field → parabolic trajectories for charged particles entering perpendicular.
• Particle 1 deflects opposite to E direction → negative charge.
• Particle 2 deflects along E direction → positive charge.
• Particle 3 deflects less than 1 but same direction as 1 → negative, but less |q/m| than 1.
Signs: 1: −, 2: +, 3: −
Highest |q/m|: Greater deflection for same speed and field implies larger |q/m|. Particle 1 deflects most → highest |q/m|.
Answer: 1: negative, 2: positive, 3: negative. Highest q/m: Particle 1.
Question 1.15
Consider a uniform electric field \(E = 3 \times 10^3 \, \text{N/C}\). (a) What is the flux through a square of side 10 cm whose plane is parallel to yz-plane? (b) What is the flux if normal to its plane makes 60° with x-axis?
Answer & Explanation:
Given: \(E = 3 \times 10^3 \, \text{N/C}\), area \(A = (0.1)^2 = 0.01 \, \text{m}^2\).
(a) Plane parallel to yz-plane:
Normal is along x-axis. If E is along x-direction, then θ = 0°.
Flux \(\phi = E A \cos\theta = 3 \times 10^3 \times 0.01 \times \cos 0^\circ = 30 \, \text{N·m}^2/\text{C}\).
(b) Normal at 60° to x-axis:
Angle between E and normal = 60°.
\(\phi = E A \cos 60^\circ = 30 \times 0.5 = 15 \, \text{N·m}^2/\text{C}\).
Answer: (a) 30 N·m²/C, (b) 15 N·m²/C.
(a) Plane parallel to yz-plane:
Normal is along x-axis. If E is along x-direction, then θ = 0°.
Flux \(\phi = E A \cos\theta = 3 \times 10^3 \times 0.01 \times \cos 0^\circ = 30 \, \text{N·m}^2/\text{C}\).
(b) Normal at 60° to x-axis:
Angle between E and normal = 60°.
\(\phi = E A \cos 60^\circ = 30 \times 0.5 = 15 \, \text{N·m}^2/\text{C}\).
Answer: (a) 30 N·m²/C, (b) 15 N·m²/C.
Question 1.16
What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so its faces are parallel to coordinate planes?
Answer & Explanation:
Key idea: Net flux through closed surface in uniform E field is zero, because equal flux enters and exits.
Explanation:
For a uniform field, every field line entering the cube exits from opposite side. Total inward flux = total outward flux.
Mathematically: \(\oint \vec{E} \cdot d\vec{A} = 0\) for uniform E over closed surface.
Answer: Net flux = 0.
Explanation:
For a uniform field, every field line entering the cube exits from opposite side. Total inward flux = total outward flux.
Mathematically: \(\oint \vec{E} \cdot d\vec{A} = 0\) for uniform E over closed surface.
Answer: Net flux = 0.
Question 1.17
Careful measurement of electric field at surface of a black box indicates net outward flux \(8.0 \times 10^5 \, \text{N·m}^2/\text{C}\). (a) What is net charge inside? (b) If flux were zero, could you conclude no charges inside?
Answer & Explanation:
(a) Using Gauss’s law: \(\phi = \frac{Q_{\text{inside}}}{\epsilon_0}\)
\(Q_{\text{inside}} = \phi \cdot \epsilon_0 = (8.0 \times 10^5) \times (8.854 \times 10^{-12})\)
≈ \(7.083 \times 10^{-6} \, \text{C} = 7.08 \, \mu\text{C}\).
(b) If flux zero:
No, zero flux only implies net charge inside is zero. There could be equal positive and negative charges inside.
Answer: (a) ~7.08 μC, (b) No, only net charge is zero.
\(Q_{\text{inside}} = \phi \cdot \epsilon_0 = (8.0 \times 10^5) \times (8.854 \times 10^{-12})\)
≈ \(7.083 \times 10^{-6} \, \text{C} = 7.08 \, \mu\text{C}\).
(b) If flux zero:
No, zero flux only implies net charge inside is zero. There could be equal positive and negative charges inside.
Answer: (a) ~7.08 μC, (b) No, only net charge is zero.
Question 1.18
A point charge +10 μC is 5 cm directly above centre of a square of side 10 cm. What is magnitude of electric flux through square? (Hint: Think of square as one face of cube with edge 10 cm.)
Answer & Explanation:
Step 1: Using symmetry
Enclose charge in a cube of side 10 cm such that charge is at centre of cube. Then flux through each face is equal by symmetry.
Total flux through cube = \(\frac{Q}{\epsilon_0}\).
\(Q = 10^{-5} \, \text{C}\), \(\epsilon_0 = 8.854 \times 10^{-12}\)
Total flux = \(\frac{10^{-5}}{8.854 \times 10^{-12}} \approx 1.129 \times 10^6 \, \text{N·m}^2/\text{C}\).
Step 2: Flux through one square face
Cube has 6 faces, so flux through one face = \(\frac{1.129 \times 10^6}{6} \approx 1.882 \times 10^5 \, \text{N·m}^2/\text{C}\).
Answer: ~\(1.88 \times 10^5 \, \text{N·m}^2/\text{C}\).
Enclose charge in a cube of side 10 cm such that charge is at centre of cube. Then flux through each face is equal by symmetry.
Total flux through cube = \(\frac{Q}{\epsilon_0}\).
\(Q = 10^{-5} \, \text{C}\), \(\epsilon_0 = 8.854 \times 10^{-12}\)
Total flux = \(\frac{10^{-5}}{8.854 \times 10^{-12}} \approx 1.129 \times 10^6 \, \text{N·m}^2/\text{C}\).
Step 2: Flux through one square face
Cube has 6 faces, so flux through one face = \(\frac{1.129 \times 10^6}{6} \approx 1.882 \times 10^5 \, \text{N·m}^2/\text{C}\).
Answer: ~\(1.88 \times 10^5 \, \text{N·m}^2/\text{C}\).
Question 1.19
A point charge of 2.0 μC is at centre of cubic Gaussian surface 9.0 cm on edge. What is net electric flux through surface?
Answer & Explanation:
By Gauss’s law: Net flux = \(\frac{Q_{\text{inside}}}{\epsilon_0}\)
\(Q = 2 \times 10^{-6} \, \text{C}\)
\(\phi = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 2.26 \times 10^5 \, \text{N·m}^2/\text{C}\).
Note: Shape/size of Gaussian surface doesn’t matter as long as it encloses charge.
Answer: \(2.26 \times 10^5 \, \text{N·m}^2/\text{C}\).
\(Q = 2 \times 10^{-6} \, \text{C}\)
\(\phi = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 2.26 \times 10^5 \, \text{N·m}^2/\text{C}\).
Note: Shape/size of Gaussian surface doesn’t matter as long as it encloses charge.
Answer: \(2.26 \times 10^5 \, \text{N·m}^2/\text{C}\).
Question 1.20
A point charge causes electric flux \(-1.0 \times 10^3 \, \text{N·m}^2/\text{C}\) through spherical Gaussian surface of 10.0 cm radius centred on charge. (a) If radius doubled, how much flux passes? (b) What is value of point charge?
Answer & Explanation:
(a) Flux independent of radius for Gaussian surface enclosing same charge. So flux remains \(-1.0 \times 10^3 \, \text{N·m}^2/\text{C}\).
(b) Using Gauss’s law: \(\phi = \frac{Q}{\epsilon_0}\)
\(Q = \phi \cdot \epsilon_0 = (-1.0 \times 10^3) \times (8.854 \times 10^{-12}) \approx -8.854 \times 10^{-9} \, \text{C}\).
Answer: (a) Same, \(-1.0 \times 10^3 \, \text{N·m}^2/\text{C}\), (b) \(-8.85 \, \text{nC}\).
(b) Using Gauss’s law: \(\phi = \frac{Q}{\epsilon_0}\)
\(Q = \phi \cdot \epsilon_0 = (-1.0 \times 10^3) \times (8.854 \times 10^{-12}) \approx -8.854 \times 10^{-9} \, \text{C}\).
Answer: (a) Same, \(-1.0 \times 10^3 \, \text{N·m}^2/\text{C}\), (b) \(-8.85 \, \text{nC}\).
Question 1.21
A conducting sphere radius 10 cm has unknown charge. Electric field 20 cm from centre is \(1.5 \times 10^3 \, \text{N/C}\) radially inward. What is net charge on sphere?
Answer & Explanation:
For conducting sphere, field outside is as if all charge at centre.
\(E = \frac{k |Q|}{r^2}\), directed inward → Q negative.
\(1.5 \times 10^3 = \frac{9 \times 10^9 |Q|}{(0.2)^2}\)
\(|Q| = \frac{1.5 \times 10^3 \times 0.04}{9 \times 10^9} = \frac{60}{9} \times 10^{-9} = 6.667 \times 10^{-9} \, \text{C}\)
So \(Q = -6.67 \times 10^{-9} \, \text{C} = -6.67 \, \text{nC}\).
Answer: \(-6.67 \, \text{nC}\).
\(E = \frac{k |Q|}{r^2}\), directed inward → Q negative.
\(1.5 \times 10^3 = \frac{9 \times 10^9 |Q|}{(0.2)^2}\)
\(|Q| = \frac{1.5 \times 10^3 \times 0.04}{9 \times 10^9} = \frac{60}{9} \times 10^{-9} = 6.667 \times 10^{-9} \, \text{C}\)
So \(Q = -6.67 \times 10^{-9} \, \text{C} = -6.67 \, \text{nC}\).
Answer: \(-6.67 \, \text{nC}\).
Question 1.22
A uniformly charged conducting sphere of 2.4 m diameter has surface charge density 80.0 μC/m². (a) Find charge on sphere. (b) Total electric flux leaving surface.
Answer & Explanation:
Radius r = 1.2 m, σ = 80 × 10⁻⁶ C/m²
(a) Charge: \(Q = \sigma \times 4\pi r^2\)
\(= 80 \times 10^{-6} \times 4 \times 3.1416 \times (1.2)^2\)
\(= 80 \times 10^{-6} \times 4 \times 3.1416 \times 1.44 \approx 1.447 \times 10^{-3} \, \text{C} = 1.447 \, \text{mC}\).
(b) Flux leaving: \(\phi = \frac{Q}{\epsilon_0} = \frac{1.447 \times 10^{-3}}{8.854 \times 10^{-12}} \approx 1.634 \times 10^8 \, \text{N·m}^2/\text{C}\).
Answer: (a) ~1.45 mC, (b) ~\(1.63 \times 10^8 \, \text{N·m}^2/\text{C}\).
(a) Charge: \(Q = \sigma \times 4\pi r^2\)
\(= 80 \times 10^{-6} \times 4 \times 3.1416 \times (1.2)^2\)
\(= 80 \times 10^{-6} \times 4 \times 3.1416 \times 1.44 \approx 1.447 \times 10^{-3} \, \text{C} = 1.447 \, \text{mC}\).
(b) Flux leaving: \(\phi = \frac{Q}{\epsilon_0} = \frac{1.447 \times 10^{-3}}{8.854 \times 10^{-12}} \approx 1.634 \times 10^8 \, \text{N·m}^2/\text{C}\).
Answer: (a) ~1.45 mC, (b) ~\(1.63 \times 10^8 \, \text{N·m}^2/\text{C}\).
Question 1.23
An infinite line charge produces field \(9 \times 10^4 \, \text{N/C}\) at distance 2 cm. Calculate linear charge density.
Answer & Explanation:
Field due to infinite line charge: \(E = \frac{\lambda}{2\pi\epsilon_0 r}\)
Given \(E = 9 \times 10^4\), \(r = 0.02 \, \text{m}\)
\(\lambda = E \cdot 2\pi\epsilon_0 r\)
\(= (9 \times 10^4) \times 2 \times 3.1416 \times 8.854 \times 10^{-12} \times 0.02\)
≈ \(9 \times 10^4 \times 1.112 \times 10^{-12} \times 0.02 \times 6.2832?\) Let’s compute:
\(2\pi\epsilon_0 r = 2\pi \times 8.854 \times 10^{-12} \times 0.02 \approx 1.112 \times 10^{-12} \times 0.12566\) Wait, compute stepwise:
\(2\pi\epsilon_0 = 5.563 \times 10^{-11}\)
Times r = 0.02 → \(1.1126 \times 10^{-12}\)
Then λ = \(9 \times 10^4 \times 1.1126 \times 10^{-12} = 1.0013 \times 10^{-7} \, \text{C/m} = 0.100 \, \mu\text{C/m}\).
Answer: ~0.100 μC/m.
Given \(E = 9 \times 10^4\), \(r = 0.02 \, \text{m}\)
\(\lambda = E \cdot 2\pi\epsilon_0 r\)
\(= (9 \times 10^4) \times 2 \times 3.1416 \times 8.854 \times 10^{-12} \times 0.02\)
≈ \(9 \times 10^4 \times 1.112 \times 10^{-12} \times 0.02 \times 6.2832?\) Let’s compute:
\(2\pi\epsilon_0 r = 2\pi \times 8.854 \times 10^{-12} \times 0.02 \approx 1.112 \times 10^{-12} \times 0.12566\) Wait, compute stepwise:
\(2\pi\epsilon_0 = 5.563 \times 10^{-11}\)
Times r = 0.02 → \(1.1126 \times 10^{-12}\)
Then λ = \(9 \times 10^4 \times 1.1126 \times 10^{-12} = 1.0013 \times 10^{-7} \, \text{C/m} = 0.100 \, \mu\text{C/m}\).
Answer: ~0.100 μC/m.
Question 1.24
Two large parallel metal plates have inner surface charge densities ±17.0 × 10⁻²² C/m². What is E: (a) outside first plate, (b) outside second plate, (c) between plates?
Answer & Explanation:
Field due to infinite sheet: \(E = \frac{\sigma}{2\epsilon_0}\) (magnitude), direction away from +, toward −.
Here σ = 17.0 × 10⁻²² C/m² for one plate, −σ for other.
Superposition:
(a) Outside first plate: fields from both plates cancel (equal and opposite) → E = 0.
(b) Outside second plate: similarly, E = 0.
(c) Between plates: fields add → \(E = \frac{\sigma}{\epsilon_0}\) from + to − plate.
\(E = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} \approx 1.92 \times 10^{-10} \, \text{N/C}\).
Answer: (a) 0, (b) 0, (c) ~\(1.92 \times 10^{-10} \, \text{N/C}\) from + to −.
Here σ = 17.0 × 10⁻²² C/m² for one plate, −σ for other.
Superposition:
(a) Outside first plate: fields from both plates cancel (equal and opposite) → E = 0.
(b) Outside second plate: similarly, E = 0.
(c) Between plates: fields add → \(E = \frac{\sigma}{\epsilon_0}\) from + to − plate.
\(E = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} \approx 1.92 \times 10^{-10} \, \text{N/C}\).
Answer: (a) 0, (b) 0, (c) ~\(1.92 \times 10^{-10} \, \text{N/C}\) from + to −.
Question 1.25
An oil drop with 12 excess electrons is stationary under constant E field 2.55 × 10⁴ N/C in Millikan’s experiment. Oil density 1.26 g/cm³. Estimate radius of drop. (g=9.81 m/s², e=1.60×10⁻¹⁹ C)
Answer & Explanation:
Equilibrium: \(qE = mg\)
\(q = 12e = 12 \times 1.6 \times 10^{-19} = 1.92 \times 10^{-18} \, \text{C}\)
\(m = \rho \times \frac{4}{3}\pi r^3\)
So: \(qE = \rho \frac{4}{3}\pi r^3 g\)
\(r^3 = \frac{3qE}{4\pi \rho g}\)
\(\rho = 1.26 \, \text{g/cm}^3 = 1260 \, \text{kg/m}^3\)
\(r^3 = \frac{3 \times 1.92 \times 10^{-18} \times 2.55 \times 10^4}{4 \times 3.1416 \times 1260 \times 9.81}\)
Numerator: \(1.46976 \times 10^{-13}\)
Denominator: \(4 \times 3.1416 \times 1260 \times 9.81 \approx 155055.5\)
\(r^3 \approx 9.48 \times 10^{-19} \, \text{m}^3\)
\(r \approx (9.48 \times 10^{-19})^{1/3} \approx 9.81 \times 10^{-7} \, \text{m} = 0.981 \, \mu\text{m}\).
Answer: ~0.98 μm.
\(q = 12e = 12 \times 1.6 \times 10^{-19} = 1.92 \times 10^{-18} \, \text{C}\)
\(m = \rho \times \frac{4}{3}\pi r^3\)
So: \(qE = \rho \frac{4}{3}\pi r^3 g\)
\(r^3 = \frac{3qE}{4\pi \rho g}\)
\(\rho = 1.26 \, \text{g/cm}^3 = 1260 \, \text{kg/m}^3\)
\(r^3 = \frac{3 \times 1.92 \times 10^{-18} \times 2.55 \times 10^4}{4 \times 3.1416 \times 1260 \times 9.81}\)
Numerator: \(1.46976 \times 10^{-13}\)
Denominator: \(4 \times 3.1416 \times 1260 \times 9.81 \approx 155055.5\)
\(r^3 \approx 9.48 \times 10^{-19} \, \text{m}^3\)
\(r \approx (9.48 \times 10^{-19})^{1/3} \approx 9.81 \times 10^{-7} \, \text{m} = 0.981 \, \mu\text{m}\).
Answer: ~0.98 μm.
Question 1.26
Which among curves in Fig. 1.35 cannot possibly represent electrostatic field lines?
Answer & Explanation:
Rules for electrostatic field lines:
1. Start on +ve charge, end on −ve charge (or infinity).
2. Never cross.
3. Density ∝ field strength.
4. No closed loops in electrostatic fields.
From figure description:
• Curves with closed loops are impossible.
• Curves crossing are impossible.
• Curves not starting/ending on charges in isolated configuration are impossible.
Likely answer: The closed loop curve and crossing curve are invalid.
1. Start on +ve charge, end on −ve charge (or infinity).
2. Never cross.
3. Density ∝ field strength.
4. No closed loops in electrostatic fields.
From figure description:
• Curves with closed loops are impossible.
• Curves crossing are impossible.
• Curves not starting/ending on charges in isolated configuration are impossible.
Likely answer: The closed loop curve and crossing curve are invalid.
Question 1.27
In a region, E is along z-direction but increases uniformly at rate 10⁵ N/C per m in +z direction. A system has total dipole moment 10⁻⁷ C·m in −z direction. Find force and torque on system.
Answer & Explanation:
Force on dipole: Non-uniform E field can exert net force.
\(F_z = p_z \frac{dE_z}{dz}\) (for small dipole).
Here \(p_z = -10^{-7} \, \text{C·m}\), \(\frac{dE_z}{dz} = 10^5 \, \text{N/C·m}\)
\(F_z = (-10^{-7}) \times 10^5 = -10^{-2} \, \text{N}\) (along −z).
Torque: \(\vec{\tau} = \vec{p} \times \vec{E}\)
Since p and E are anti-parallel (p along −z, E along +z), τ = 0.
Answer: Force = −0.01 N along z, Torque = 0.
\(F_z = p_z \frac{dE_z}{dz}\) (for small dipole).
Here \(p_z = -10^{-7} \, \text{C·m}\), \(\frac{dE_z}{dz} = 10^5 \, \text{N/C·m}\)
\(F_z = (-10^{-7}) \times 10^5 = -10^{-2} \, \text{N}\) (along −z).
Torque: \(\vec{\tau} = \vec{p} \times \vec{E}\)
Since p and E are anti-parallel (p along −z, E along +z), τ = 0.
Answer: Force = −0.01 N along z, Torque = 0.
Question 1.28
(a) Conductor A with cavity given charge Q. Show all charge appears on outer surface. (b) Conductor B with charge q inserted into cavity (insulated). Show outer surface charge of A becomes Q+q. (c) Suggest shielding method for sensitive instrument.
Answer & Explanation:
(a) Inside conductor, E=0. By Gauss’s law, net charge inside any Gaussian surface within conductor is zero. So induced charge on cavity wall cancels any internal charge. Given only Q on conductor, it resides on outer surface.
(b) When B with q is inserted, charge −q is induced on cavity wall of A (by electrostatic induction). To keep net charge of A as Q, outer surface must have Q + q.
(c) Enclose instrument in a conducting shield (Faraday cage). External fields cause rearrangement of charges on shield, making interior field zero.
(b) When B with q is inserted, charge −q is induced on cavity wall of A (by electrostatic induction). To keep net charge of A as Q, outer surface must have Q + q.
(c) Enclose instrument in a conducting shield (Faraday cage). External fields cause rearrangement of charges on shield, making interior field zero.
Question 1.29
A hollow charged conductor has tiny hole cut into surface. Show electric field in hole is \((\sigma / 2\epsilon_0) \hat{n}\), where σ is surface charge density near hole, \(\hat{n}\) outward normal.
Answer & Explanation:
Approach: Superposition principle.
Consider hole as small removed disk. Field just outside conductor (without hole) is \(\frac{\sigma}{\epsilon_0} \hat{n}\).
Removing disk is equivalent to superimposing a disk of charge −σ on that spot.
Field due to a charged disk (very small) near its centre is \(\frac{\sigma}{2\epsilon_0}\) normally away from disk.
For −σ disk, field is \(-\frac{\sigma}{2\epsilon_0} \hat{n}\) inside conductor.
Superpose: Original field at that point (just outside) was \(\frac{\sigma}{\epsilon_0} \hat{n}\). After removing disk, net field in hole = \(\frac{\sigma}{\epsilon_0} \hat{n} - \frac{\sigma}{2\epsilon_0} \hat{n} = \frac{\sigma}{2\epsilon_0} \hat{n}\).
This is the field just inside the hole.
Consider hole as small removed disk. Field just outside conductor (without hole) is \(\frac{\sigma}{\epsilon_0} \hat{n}\).
Removing disk is equivalent to superimposing a disk of charge −σ on that spot.
Field due to a charged disk (very small) near its centre is \(\frac{\sigma}{2\epsilon_0}\) normally away from disk.
For −σ disk, field is \(-\frac{\sigma}{2\epsilon_0} \hat{n}\) inside conductor.
Superpose: Original field at that point (just outside) was \(\frac{\sigma}{\epsilon_0} \hat{n}\). After removing disk, net field in hole = \(\frac{\sigma}{\epsilon_0} \hat{n} - \frac{\sigma}{2\epsilon_0} \hat{n} = \frac{\sigma}{2\epsilon_0} \hat{n}\).
This is the field just inside the hole.
Question 1.30
Obtain formula for electric field due to long thin wire of uniform λ without Gauss’s law. [Hint: Use Coulomb’s law and integrate.]
Answer & Explanation:
Set up: Wire along z-axis, point P at perpendicular distance r in x-y plane.
Element dz at coordinate z has charge λ dz.
Distance from dz to P: \(R = \sqrt{r^2 + z^2}\).
Field dE at P: magnitude \(\frac{1}{4\pi\epsilon_0} \frac{\lambda dz}{R^2}\).
By symmetry, horizontal components cancel, vertical (radial) components add.
Radial component: \(dE_r = dE \cos\theta = dE \cdot \frac{r}{R}\).
Integrate over z from −∞ to +∞:
\(E = \int dE_r = \frac{\lambda r}{4\pi\epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{(r^2+z^2)^{3/2}}\).
Integral = \(\frac{2}{r^2}\). So \(E = \frac{\lambda}{2\pi\epsilon_0 r}\).
Result: \(E = \frac{\lambda}{2\pi\epsilon_0 r}\) radially outward (if λ positive).
Element dz at coordinate z has charge λ dz.
Distance from dz to P: \(R = \sqrt{r^2 + z^2}\).
Field dE at P: magnitude \(\frac{1}{4\pi\epsilon_0} \frac{\lambda dz}{R^2}\).
By symmetry, horizontal components cancel, vertical (radial) components add.
Radial component: \(dE_r = dE \cos\theta = dE \cdot \frac{r}{R}\).
Integrate over z from −∞ to +∞:
\(E = \int dE_r = \frac{\lambda r}{4\pi\epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{(r^2+z^2)^{3/2}}\).
Integral = \(\frac{2}{r^2}\). So \(E = \frac{\lambda}{2\pi\epsilon_0 r}\).
Result: \(E = \frac{\lambda}{2\pi\epsilon_0 r}\) radially outward (if λ positive).
Question 1.31
Protons and neutrons consist of three quarks each. Up quark u has charge +2e/3, down quark d has charge −e/3. Suggest quark composition of proton and neutron.
Answer & Explanation:
Proton charge = +e: Possible combination: uud → total charge = \(2/3 + 2/3 - 1/3 = 3/3 = +e\).
Neutron charge = 0: Possible: udd → \(2/3 - 1/3 - 1/3 = 0\).
Answer: Proton: uud, Neutron: udd.
Neutron charge = 0: Possible: udd → \(2/3 - 1/3 - 1/3 = 0\).
Answer: Proton: uud, Neutron: udd.
Question 1.32
(a) Show equilibrium of test charge at null point (E=0) is necessarily unstable. (b) Verify for two equal same-sign charges.
Answer & Explanation:
(a) General proof:
At null point, net force zero. If displaced slightly, field lines converge or diverge. For electrostatic field, field lines originate/terminate on charges, so null point is saddle point → restoring force not in all directions → unstable equilibrium.
(b) Example: Two equal positive charges separated. Midpoint has E=0. Displace along line joining charges → net force away from midpoint. Displace perpendicular → net force toward midpoint? Wait, actually along perpendicular, symmetry gives restoring? Let’s check: For two same charges, midpoint is stable for perpendicular displacement? Actually, potential has minimum in perpendicular direction? Known: Two like charges: null point at midpoint is unstable along line, stable perpendicular? But overall, since unstable in at least one direction, equilibrium is unstable.
At null point, net force zero. If displaced slightly, field lines converge or diverge. For electrostatic field, field lines originate/terminate on charges, so null point is saddle point → restoring force not in all directions → unstable equilibrium.
(b) Example: Two equal positive charges separated. Midpoint has E=0. Displace along line joining charges → net force away from midpoint. Displace perpendicular → net force toward midpoint? Wait, actually along perpendicular, symmetry gives restoring? Let’s check: For two same charges, midpoint is stable for perpendicular displacement? Actually, potential has minimum in perpendicular direction? Known: Two like charges: null point at midpoint is unstable along line, stable perpendicular? But overall, since unstable in at least one direction, equilibrium is unstable.
Question 1.33
Particle mass m, charge −q enters region between charged plates moving along x with speed v_x. Plate length L, uniform E maintained. Show vertical deflection at far edge is \(qEL^2 / (2m v_x^2)\). Compare with projectile motion.
Answer & Explanation:
Motion analysis:
Horizontal: uniform motion, time to cross plates \(t = L / v_x\).
Vertical: constant acceleration \(a = F/m = qE/m\) (downward if −q, E upward, but sign handled).
Deflection \(y = \frac{1}{2} a t^2 = \frac{1}{2} \frac{qE}{m} \left( \frac{L}{v_x} \right)^2 = \frac{qEL^2}{2m v_x^2}\).
Comparison with projectile: Identical to motion under gravity with g replaced by \(qE/m\). Horizontal uniform motion, vertical uniformly accelerated.
Horizontal: uniform motion, time to cross plates \(t = L / v_x\).
Vertical: constant acceleration \(a = F/m = qE/m\) (downward if −q, E upward, but sign handled).
Deflection \(y = \frac{1}{2} a t^2 = \frac{1}{2} \frac{qE}{m} \left( \frac{L}{v_x} \right)^2 = \frac{qEL^2}{2m v_x^2}\).
Comparison with projectile: Identical to motion under gravity with g replaced by \(qE/m\). Horizontal uniform motion, vertical uniformly accelerated.
Question 1.34
Suppose particle in 1.33 is electron with \(v_x = 2.0 \times 10^6 \, \text{m/s}\). If E = 9.1 × 10² N/C between plates separated by 0.5 cm, where will electron strike upper plate? (e=1.6×10⁻¹⁹ C, m_e=9.1×10⁻³¹ kg)
Answer & Explanation:
Given: d = 0.005 m (plate separation), E upward. Electron (−q) enters midway? Usually enters at lower plate? Assume starts at lower plate, strikes upper.
Need horizontal distance L from entry point to strike point.
Vertical motion: \(y = d = \frac{1}{2} a t^2\), \(a = \frac{eE}{m}\) (upward force on electron is opposite E, so downward? Wait: electron charge −e, E upward → force downward. So electron accelerates downward, but if starts at lower plate, it won’t reach upper. So must start at upper plate? Or E downward? Clarify: In problem, likely E is such that electron deflected upward. Let’s assume E downward so force on electron upward. Set a = eE/m upward.
Then d = \(\frac{1}{2} a t^2\) → \(t = \sqrt{2d / a}\).
Horizontal distance travelled in this time: \(L = v_x t = v_x \sqrt{\frac{2d m}{eE}}\).
Compute: \(a = \frac{1.6 \times 10^{-19} \times 9.1 \times 10^2}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2\)
\(t = \sqrt{\frac{2 \times 0.005}{1.6 \times 10^{14}}} = \sqrt{6.25 \times 10^{-17}} = 7.906 \times 10^{-9} \, \text{s}\)
\(L = 2.0 \times 10^6 \times 7.906 \times 10^{-9} = 0.0158 \, \text{m} = 1.58 \, \text{cm}\).
Answer: Strikes upper plate 1.58 cm from entry point.
Need horizontal distance L from entry point to strike point.
Vertical motion: \(y = d = \frac{1}{2} a t^2\), \(a = \frac{eE}{m}\) (upward force on electron is opposite E, so downward? Wait: electron charge −e, E upward → force downward. So electron accelerates downward, but if starts at lower plate, it won’t reach upper. So must start at upper plate? Or E downward? Clarify: In problem, likely E is such that electron deflected upward. Let’s assume E downward so force on electron upward. Set a = eE/m upward.
Then d = \(\frac{1}{2} a t^2\) → \(t = \sqrt{2d / a}\).
Horizontal distance travelled in this time: \(L = v_x t = v_x \sqrt{\frac{2d m}{eE}}\).
Compute: \(a = \frac{1.6 \times 10^{-19} \times 9.1 \times 10^2}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2\)
\(t = \sqrt{\frac{2 \times 0.005}{1.6 \times 10^{14}}} = \sqrt{6.25 \times 10^{-17}} = 7.906 \times 10^{-9} \, \text{s}\)
\(L = 2.0 \times 10^6 \times 7.906 \times 10^{-9} = 0.0158 \, \text{m} = 1.58 \, \text{cm}\).
Answer: Strikes upper plate 1.58 cm from entry point.
📘 Exam Preparation Tip:
These exercise questions will help you understand the fundamental principles of electrostatics. You'll learn to apply Coulomb's law to calculate forces between charges, determine electric fields due to point charges and continuous distributions, use Gauss's law to find fields for symmetric charge configurations, and analyze dipole fields and torque. Essential for building a strong foundation in electromagnetism and solving complex field problems in board exams and competitive tests.