JEE Physics Electric Charges And Fields MCQs Set 29

Multiple Choice Questions (MCQs) for JEE Physics: Electric Charges and Fields

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Multiple Choice Questions

Question 1: A hollow insulated conduction sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere if its radius is 2 m?
(a) zero
(b) \( 5\ \mu\text{C m}^{-2} \)
(c) \( 20\ \mu\text{C m}^{-2} \)
(d) \( 8\ \mu\text{C m}^{-2} \)
Show Answer & Explanation

Answer: (a) zero

Explanation:
1. The charge resides on the outer surface of the conductor.
2. The field inside a hollow conductor, including at its centre, is zero.

Teacher's Note:
1. Field inside a conductor is zero.

Question 2: A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
(a) increase four times
(b) be reduced to half
(c) remain the same
(d) be doubled
Show Answer & Explanation

Answer: (c) remain the same

Explanation:
1. By Gauss's law the flux depends only on the enclosed charge, Q/ε₀.
2. It does not depend on the radius.

Teacher's Note:
1. Φ = Q/ε₀.

Question 3: If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium, then the value of q is
(a) \( Q/2 \)
(b) \( -Q/2 \)
(c) \( Q/4 \)
(d) \( -Q/4 \)
Show Answer & Explanation

Answer: (d) \( -Q/4 \)

Explanation:
1. For equilibrium of one charge Q (separation d): kQ²/d² + kQq/(d/2)² = 0.
2. So Q + 4q = 0 and q = −Q/4.

Teacher's Note:
1. Balance the forces on an end charge.

Question 4: Two charges q and −3q are placed fixed on the x-axis separated by a distance d (q at A, −3q at B). At what distance from A, on the side away from B, should a third charge 2q be placed so that it does not experience any force?
(a) \( \frac{d-\sqrt3d}{2} \)
(b) \( \frac{d+\sqrt3d}{2} \)
(c) \( \frac{d+3d}{2} \)
(d) \( \frac{d-3d}{2} \)
Show Answer & Explanation

Answer: (b) \( \frac{d+\sqrt3d}{2} \)

Explanation:
1. Let the distance from A be x on the far side from B.
2. Zero force needs kq/x² = 3kq/(x + d)², so x + d = √3 x and x = d/(√3 − 1) = d(√3 + 1)/2.

Teacher's Note:
1. Equate the two forces.

Question 5: A rod of length 2.4 m and radius 4.6 mm carries a negative charge of 4.2 × 10⁻⁷ C spread uniformly over its surface. The electric field near the mid-point of the rod, at a point on its surface, is
(a) \( -8.6\times10^5\ \text{N C}^{-1} \)
(b) \( 8.6\times10^4\ \text{N C}^{-1} \)
(c) \( -6.7\times10^5\ \text{N C}^{-1} \)
(d) \( 6.7\times10^4\ \text{N C}^{-1} \)
Show Answer & Explanation

Answer: (c) \( -6.7\times10^5\ \text{N C}^{-1} \)

Explanation:
1. λ = 4.2 × 10⁻⁷/2.4 = 1.75 × 10⁻⁷ C/m, and E = λ/(2πε₀r) = 2 × 9 × 10⁹ × 1.75 × 10⁻⁷/4.6 × 10⁻³ ≈ 6.8 × 10⁵ N/C.
2. For a negative charge the field points inward, giving about −6.7 × 10⁵ N/C.

Teacher's Note:
1. Long-rod formula.

Question 6:
Assertion (A): On disturbing an electric dipole in stable equilibrium in an electric field, it returns to its stable equilibrium orientation.
Reason (R): A restoring torque acts on the dipole when it is disturbed from stable equilibrium.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The torque τ = p × E turns the dipole back towards the field direction.
2. So the restoring torque is the reason for the return.

Teacher's Note:
1. Restoring torque.

Question 7:
Assertion (A): The Coulomb force and the gravitational force follow the same inverse-square law.
Reason (R): Both laws are the same in all aspects.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Both forces vary as 1/r², so the assertion is true.
2. The Coulomb force can be attractive or repulsive and is far stronger, while gravity is always attractive, so the reason is false.

Teacher's Note:
1. Similar form, different nature.

Question 8: Select the correct statements. I. Inside a charged or neutral conductor, the electrostatic field is zero. II. The electrostatic field at the surface of a charged conductor must be tangential to the surface at any point. III. There is no net charge at any point inside the conductor.
(a) I and II
(b) I and III
(c) II and III
(d) I, II and III
Show Answer & Explanation

Answer: (b) I and III

Explanation:
1. In electrostatics the field inside a conductor is zero and there is no net charge inside.
2. The field at the surface is normal, not tangential, so II is false.

Teacher's Note:
1. Surface field is normal.

Question 9: In a uniform electric field E, a charge +q of negligible mass is released at a point. Which of the following statements are correct? I. Its velocity increases with time. II. A force acts on it in the direction of the electric field. III. Its momentum changes with time.
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Show Answer & Explanation

Answer: (d) I, II and III

Explanation:
1. The force qE acts on the charge along the field.
2. It gives a constant acceleration, so the velocity and momentum both change with time.

Teacher's Note:
1. F = qE.

Question 10: On rubbing, when one body gets positively charged and the other negatively charged, the electrons transferred from the positively charged body to the negatively charged body are
(a) valence electrons only
(b) electrons of inner shells
(c) both valence electrons and electrons of the inner shell
(d) yet to be established
Show Answer & Explanation

Answer: (a) valence electrons only

Explanation:
1. Only the loosely bound outermost (valence) electrons are transferred during rubbing.

Teacher's Note:
1. Valence electrons move.

Question 11:
Assertion (A): A metallic shield in the form of a hollow shell may be built to block an electric field.
Reason (R): In a hollow spherical shield the electric field inside it is zero at every point.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The field inside a conducting cavity is zero, so the region is shielded.
2. The reason explains the assertion.

Teacher's Note:
1. Electrostatic shielding.

Question 12: Two particles of equal mass m and charge q are placed at a distance of 16 cm. They do not experience any force. The value of q/m is
(a) 1
(b) \( \sqrt{\frac{\pi\varepsilon_0}{G}} \)
(c) \( \sqrt{\frac{G}{4\pi\varepsilon_0}} \)
(d) \( \sqrt{4\pi\varepsilon_0G} \)
Show Answer & Explanation

Answer: (d) \( \sqrt{4\pi\varepsilon_0G} \)

Explanation:
1. No net force means q²/(4πε₀r²) = Gm²/r².
2. So (q/m)² = 4πε₀G.

Teacher's Note:
1. Coulomb force = gravitational force.

Question 13: The number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is
(a) \( 1.13\times10^{11} \)
(b) \( 1.13\times10^{10} \)
(c) \( 0.61\times10^{11} \)
(d) \( 0.61\times10^{9} \)
Show Answer & Explanation

Answer: (a) \( 1.13\times10^{11} \)

Explanation:
1. The number of lines equals the flux q/ε₀.
2. = 1/(8.85 × 10⁻¹²) = 1.13 × 10¹¹.

Teacher's Note:
1. N = q/ε₀.

Question 14: Two point charges +Q and +q are separated by a certain distance. If Q > q, then in between the charges the electric field is zero at a point
(a) closer to +Q
(b) closer to +q
(c) exactly at the mid-point of the line segment joining +Q and +q
(d) nowhere on the line segment joining +Q and +q
Show Answer & Explanation

Answer: (b) closer to +q

Explanation:
1. The fields cancel where kQ/x₁² = kq/x₂².
2. Since Q > q, this needs x₁ > x₂, so the point is closer to the smaller charge +q.

Teacher's Note:
1. Null point is nearer the smaller charge.

Question 15: Two insulated charged metallic spheres P and Q have their centres separated by a distance of 60 cm. The radii of P and Q are negligible compared to the distance of separation. The mutual force of electrostatic repulsion if the charge on each is 3.2 × 10⁻⁷ C is
(a) \( 5.2\times10^{-4}\ \text{N} \)
(b) \( 2.5\times10^{-3}\ \text{N} \)
(c) \( 1.5\times10^{-3}\ \text{N} \)
(d) \( 3.5\times10^{-4}\ \text{N} \)
Show Answer & Explanation

Answer: (b) \( 2.5\times10^{-3}\ \text{N} \)

Explanation:
1. F = 9 × 10⁹ × (3.2 × 10⁻⁷)²/(0.6)².
2. = 9.216 × 10⁻⁴/0.36 ≈ 2.56 × 10⁻³ ≈ 2.5 × 10⁻³ N.

Teacher's Note:
1. F = kq²/r².

Question 16: For distances far away from the centre of a dipole, the change in magnitude of the electric field with change in distance from the centre of the dipole is
(a) zero
(b) the same in the equatorial plane as well as on the axis of the dipole
(c) more in the case of the equatorial plane of the dipole as compared to its axis
(d) more in the case of the axis of the dipole as compared to the equatorial plane
Show Answer & Explanation

Answer: (d) more in the case of the axis of the dipole as compared to the equatorial plane

Explanation:
1. E_axis = 2kp/r³ and E_equator = kp/r³, both falling as 1/r³.
2. The rate of change is twice as large on the axis.

Teacher's Note:
1. dE/dr is larger when E is larger.

Question 17: Three charges q, Q and 4q are placed in a straight line of length l at points distant 0, l/2 and l respectively from one end. In order to make the net force on q zero, the charge Q must be equal to
(a) \( -q \)
(b) \( -2q \)
(c) \( \frac{-q}{2} \)
(d) \( q \)
Show Answer & Explanation

Answer: (a) \( -q \)

Explanation:
1. The force on q from 4q is kq·4q/l² (repulsive).
2. The force from Q must cancel it: kqQ/(l/2)² = 4kqQ/l² with the opposite sense, so Q = −q.

Teacher's Note:
1. Balance forces on q.

Question 18: The magnitude of the electric field just sufficient to balance the earth's gravitational attraction on an electron is (mass of electron 9.1 × 10⁻³¹ kg, charge 1.6 × 10⁻¹⁹ C, g = 9.8 m/s²)
(a) \( 5.6\times10^{-11}\ \text{N/C} \)
(b) \( 4.8\times10^{-15}\ \text{N/C} \)
(c) \( 1.6\times10^{-19}\ \text{N/C} \)
(d) \( 3.2\times10^{-19}\ \text{N/C} \)
Show Answer & Explanation

Answer: (a) \( 5.6\times10^{-11}\ \text{N/C} \)

Explanation:
1. eE = mg.
2. E = 9.1 × 10⁻³¹ × 9.8/(1.6 × 10⁻¹⁹) = 5.6 × 10⁻¹¹ N/C.

Teacher's Note:
1. Set eE = mg.

Question 19: A body is positively charged. This implies that
(a) there is only positive charge in the body
(b) there is positive as well as negative charge in the body but the positive charge is more than the negative charge
(c) there is equal positive and negative charge in the body but the positive charge lies in the outer regions
(d) negative charge is displaced from its position
Show Answer & Explanation

Answer: (b) there is positive as well as negative charge in the body but the positive charge is more than the negative charge

Explanation:
1. A body contains equal numbers of protons and electrons when neutral.
2. It is positively charged when it has lost some electrons, so the positive charge exceeds the negative charge.

Teacher's Note:
1. Positive charging means electron deficit.

Question 20: Let E_a be the electric field strength of a short dipole at a point on its axial line and E_e that on the equatorial line at the same distance. Then
(a) \( E_e=2E_a \)
(b) \( E_a=2E_e \)
(c) \( E_a=E_e \)
(d) none of the above
Show Answer & Explanation

Answer: (b) \( E_a=2E_e \)

Explanation:
1. E_axial = 2kp/r³ and E_equatorial = kp/r³.
2. So E_a = 2E_e.

Teacher's Note:
1. Axis field is double.

Question 21:
Assertion (A): If there exists a Coulomb attraction between two bodies, both of them may not be charged.
Reason (R): In Coulomb attraction the two bodies are oppositely charged.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. A charged body also attracts a neutral body by induction, so both need not be charged.
2. The reason says both are charged, so it is false.

Teacher's Note:
1. Induction attracts neutral bodies.

Question 22: Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between them, the number of electrons missing from each ion will be (e being the charge of an electron)
(a) \( \frac{4\pi\varepsilon_0Fd^2}{e^2} \)
(b) \( \sqrt{\frac{4\pi\varepsilon_0Fe^2}{d^2}} \)
(c) \( \sqrt{\frac{4\pi\varepsilon_0Fd^2}{e^2}} \)
(d) \( \frac{4\pi\varepsilon_0Fd^2}{q^2} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{\frac{4\pi\varepsilon_0Fd^2}{e^2}} \)

Explanation:
1. F = q²/(4πε₀d²), so q = d√(4πε₀F).
2. The number of missing electrons is n = q/e = √(4πε₀Fd²/e²).

Teacher's Note:
1. n = q/e.

Question 23: What happens when some charge is placed on a soap bubble?
(a) its radius decreases
(b) its radius increases
(c) the bubble collapses
(d) none of these
Show Answer & Explanation

Answer: (b) its radius increases

Explanation:
1. The like charges on the bubble repel each other and push the surface outward.
2. So the radius increases.

Teacher's Note:
1. Electrostatic pressure expands the bubble.

Question 24:
Assertion (A): The flux over a given Gaussian surface depends only on the charges enclosed (q₁ and q₂) by the surface.
Reason (R): The electric field at all points on the Gaussian surface depends only on charges q₁ and q₂.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Gauss's law gives flux from the enclosed charge only, so the assertion is true.
2. The field at points of the surface depends on all charges, inside and outside, so the reason is false.

Teacher's Note:
1. Field depends on every charge.

Question 25: An electric dipole is placed at the centre of a sphere. Then
(a) the flux of the electric field through the sphere is not zero
(b) the electric field is zero at every point of the sphere
(c) the electric field is not zero anywhere on the sphere
(d) the electric field is zero on the sphere
Show Answer & Explanation

Answer: (c) the electric field is not zero anywhere on the sphere

Explanation:
1. The net charge of a dipole is zero, so the net flux is zero.
2. The field of a dipole is non-zero at every finite point, so it is not zero anywhere on the sphere.

Teacher's Note:
1. Zero net flux does not mean zero field.

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