Electric Charges and Coulomb's Law – Electric charges: NEET MCQs with Answers

15 free questions on Electric Charges and Coulomb's Law – Electric charges for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Electric charges: Conservation of charge. Coulomb's law forces between two point charges, forces between multiple charges: superposition principle

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Question 1

Previous-year question
Two identical charged spheres suspended from a common point by two massless strings of length l are initially at a distance d (d ≪ l) apart because of their mutual repulsion. The charges begin to leak from both spheres at a constant rate, and as a result the spheres approach each other with a velocity v. Then v varies as a function of the distance x between them as [NEET Phase I 2016]
  1. v ∝ x⁻¹ᐟ²
  2. v ∝ x⁻¹
  3. v ∝ x¹ᐟ²
  4. v ∝ x
Show answer
Answer: (A) v ∝ x⁻¹ᐟ²
Explanation: For the suspended sphere, T cos θ = mg and T sin θ = kq²/x². Dividing, tan θ = kq²/(x²mg), and for small θ, tan θ ≈ sin θ = x/2l. This gives q² ∝ x³, i.e. q ∝ x^(3/2). Differentiating, dq/dt ∝ x^(1/2)(dx/dt) = x^(1/2)v. Since the leakage rate dq/dt is constant, v ∝ x⁻¹ᐟ².

Question 2

Previous-year question
Two point charges A and B, having charges +Q and −Q respectively, are placed a certain distance apart and the force acting between them is F. If 25 % of the charge of A is transferred to B, then the force between the charges becomes [NEET 2019]
  1. F
  2. 9F/16
  3. 16F/9
  4. 4F/3
Show answer
Answer: (B) 9F/16
Explanation: After the transfer A carries 0.75Q and B carries −Q + 0.25Q = −0.75Q. Since F ∝ q₁q₂, the new force is F × (0.75)(0.75) = 9F/16.

Question 3

Previous-year question
Suppose the charge of a proton and an electron differ slightly. One of them is −e, the other is (e + Δe). If the net of the electrostatic force and the gravitational force between two hydrogen atoms placed at a distance d apart is zero, then Δe is of the order of (mass of hydrogen m_h = 1.67 × 10⁻²⁷ kg) [NEET 2017]
  1. 10⁻²³ C
  2. 10⁻³⁷ C
  3. 10⁻⁴⁷ C
  4. 10⁻²⁰ C
Show answer
Answer: (B) 10⁻³⁷ C
Explanation: Setting k(Δe)²/d² = Gm_h²/d² gives Δe = m_h√(G/k) = 1.67 × 10⁻²⁷ × √(6.67 × 10⁻¹¹/9 × 10⁹) ≈ 1.4 × 10⁻³⁷ C.

Question 4

Previous-year question
Two identical charged spheres suspended from a common point by two massless strings of length l are initially at a distance d (d ≪ l) apart because of their mutual repulsion. The charges begin to leak from both spheres at a constant rate. As a result the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres as [NEET 2016]
  1. v ∝ x^(1/2)
  2. v ∝ x
  3. v ∝ x^(−1/2)
  4. v ∝ x^(−1)
Show answer
Answer: (C) v ∝ x^(−1/2)
Explanation: Equilibrium gives x³ ∝ q², and since the charge leaks at a constant rate q ∝ t, we get x ∝ t^(2/3). Differentiating, v ∝ t^(−1/3) ∝ x^(−1/2).

Question 5

Point charges q₁ = −5 μC and q₂ = +10 μC are fixed on the y-axis at (0, +1 m) and (0, −1 m) respectively. A small object of charge +4 μC and mass 4.5 g is released from rest at the point P (√3 m, 0). Ignoring gravity, its initial acceleration in m s⁻² is (take 1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. 5√3 î + 5 ĵ
  2. 5√3 î + 15 ĵ
  3. 15 î + 5√3 ĵ
  4. 5 î + 5√3 ĵ
Show answer
Answer: (B) 5√3 î + 15 ĵ
Explanation: Each fixed charge is 2 m from P. The repulsion from q₂ is 0.09 N, giving 20 m s⁻² along (√3/2, 1/2), that is (10√3, 10). The attraction towards q₁ is 0.045 N, giving 10 m s⁻² along (−√3/2, 1/2), that is (−5√3, 5). Adding the components gives 5√3 î + 15 ĵ.

Question 6

A, B, C and D are consecutive corners of a square whose centre is O. Charges +2q, +3q, +2q and −3q are fixed at A, B, C and D respectively. What is the direction of the force on a small positive charge placed at O?
  1. Along the diagonal AC
  2. The force is zero
  3. Along the diagonal BD, directed from D towards B
  4. Along the diagonal BD, directed from B towards D
Show answer
Answer: (D) Along the diagonal BD, directed from B towards D
Explanation: A and C are opposite corners carrying equal charges at equal distances from O, so their forces cancel. The positive charge at O is repelled by +3q at B, which pushes it towards D, and attracted by −3q at D, which also pulls it towards D. These two forces add, so the net force points along BD from B towards D.

Question 7

Electric charge is often compared with mass. Which one of the following statements is NOT correct?
  1. Like mass, charge has magnitude but no direction
  2. Mass obeys a basic law of quantisation in integral multiples of a smallest unit, exactly as charge does
  3. The mass of a body is always positive, whereas a charge can be positive or negative
  4. The total charge of a system of +5 μC, −8 μC, +2 μC and +3 μC is +2 μC
Show answer
Answer: (B) Mass obeys a basic law of quantisation in integral multiples of a smallest unit, exactly as charge does
Explanation: Charge and mass are both scalars, but charges must be added with their proper signs, so 5 − 8 + 2 + 3 gives +2 μC. Mass is always positive, while charge can have either sign. Quantisation of charge in units of e is a basic law of nature, and there is no analogous law for mass.
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Question 8

Point charges +9 μC and −4 μC are fixed at x = 0 and x = 10 cm respectively. A third point charge Q is to be placed on the x-axis so that the net force on each of the three charges is zero. What are the value of Q and its position?
  1. +1.44 μC at x = 6 cm
  2. −36 μC at x = −20 cm
  3. −36 μC at x = 30 cm
  4. +36 μC at x = 30 cm
Show answer
Answer: (D) +36 μC at x = 30 cm
Explanation: Between unlike charges the two forces on Q add, so Q must lie outside the pair, on the side of the smaller charge. If it is y beyond x = 10 cm, 9/(10 + y)² = 4/y² gives y = 20 cm, so x = 30 cm. The −4 μC charge is pulled towards the origin by +9 μC, so Q must attract it the other way and must be positive. Then 9 × 4/10² = 4Q/20² gives Q = +36 μC, and the force on +9 μC also balances.

Question 9

A small metal ball carries a charge of 1.6 μC (e = 1.6 × 10⁻¹⁹ C). Charge is quantised, yet in calculations this charge is treated as able to vary continuously. Why is this acceptable?
  1. The ball holds about 10¹⁹ excess elementary charges, and quantisation applies only to single atoms
  2. Quantisation of charge holds only for insulators, so it does not apply to a metal ball
  3. The excess charge on a metal spreads over the surface, and spreading removes the quantisation
  4. The ball holds about 10¹³ excess elementary charges, so a change of one e is too small a fraction to detect
Show answer
Answer: (D) The ball holds about 10¹³ excess elementary charges, so a change of one e is too small a fraction to detect
Explanation: The number of elementary charges is n = q/e = 1.6 × 10⁻⁶ / 1.6 × 10⁻¹⁹ = 10¹³. Adding or removing one elementary charge changes the total by only about 1 part in 10¹³. At the macroscopic level the steps are too small to notice, so charge behaves as if continuous, though it is still quantised.

Question 10

An isolated neutron changes into a proton and an electron. A student says that charge has been created here, so conservation of charge fails. What is the correct conclusion?
  1. The law is obeyed because it fixes the number of charged particles, which does not change here
  2. Charge is conserved because the proton's charge is slightly smaller than the electron's charge
  3. The law is obeyed because the new charges are equal and opposite, so the net charge stays zero
  4. Conservation of charge fails, but only for processes inside the nucleus
Show answer
Answer: (C) The law is obeyed because the new charges are equal and opposite, so the net charge stays zero
Explanation: Conservation of charge says that the total charge of an isolated system remains constant. It does not say that the number of charged particles is fixed. The neutron has zero charge, and the proton and electron carry +e and −e, so the net charge after the change is also zero.

Question 11

Electric charge and mass are both scalar quantities that add up for a body. Which statement correctly brings out the difference in how they add?
  1. Charges add algebraically only in conductors, while in insulators the total charge cannot be found by simple addition
  2. Mass can be positive or negative, but charge is always positive, so only the total mass of a body can be zero
  3. Charges add algebraically with their signs, while mass is always positive, so a body with many charges can have zero net charge but never zero mass
  4. Charges add vectorially while masses add algebraically, so the direction of each charge matters
Show answer
Answer: (C) Charges add algebraically with their signs, while mass is always positive, so a body with many charges can have zero net charge but never zero mass
Explanation: Charge is a scalar and has a sign, so the total charge of a system is the algebraic sum of its individual charges. Positive and negative charges can therefore cancel completely. Mass has no such sign, so the masses of the parts always add to a positive total.

Question 12

Two fixed point charges, each +Q, are placed at x = −a and x = +a. A particle of mass m and charge −q is at the origin. It is displaced a small distance y (y ≪ a) along the y-axis and released. Neglecting gravity, how does the particle behave?
  1. It oscillates with period 2π√(2πε₀ma³/Qq)
  2. It oscillates with period 2π√(πε₀ma³/Qq)
  3. It oscillates with period 2π√(4πε₀ma³/Qq)
  4. It moves steadily away from the origin along the y-axis
Show answer
Answer: (A) It oscillates with period 2π√(2πε₀ma³/Qq)
Explanation: Each charge attracts the particle with force Qq/[4πε₀(a² + y²)], and the x-components cancel. The y-components add to give a force 2 × Qq/(4πε₀a²) × (y/a) = Qqy/(2πε₀a³) towards the origin, for y ≪ a. This is a restoring force proportional to y, so the motion is simple harmonic with ω² = Qq/(2πε₀ma³). The period is then 2π√(2πε₀ma³/Qq).

Question 13

Two small blocks, each of mass 40 g, rest on a rough horizontal table with their centres 30 cm apart. The coefficient of static friction is 0.25. One block carries a charge q and the other a charge 4q. What is the largest q for which both blocks stay at rest? (g = 10 m s⁻², 1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. 0.25 μC
  2. 0.50 μC
  3. 1.0 μC
  4. 2.0 μC
Show answer
Answer: (B) 0.50 μC
Explanation: By Newton's third law, the electric force on each block is the same, even though the charges differ. Limiting friction on each block is 0.25 × 0.040 × 10 = 0.10 N. Setting (9 × 10⁹)(4q²)/(0.30)² = 0.10 gives q² = 2.5 × 10⁻¹³ C². Hence q = 0.50 μC.

Question 14

Two identical small conducting spheres carry charges of opposite sign and unequal magnitude, and attract each other with a force F. They are touched together and returned to their original positions, after which they repel each other with a force 16F/9. The ratio of the larger to the smaller initial charge magnitude is
  1. 9 : 1
  2. 4 : 1
  3. 3 : 1
  4. 16 : 9
Show answer
Answer: (A) 9 : 1
Explanation: Let the magnitudes be nq and q. After contact each sphere carries (n − 1)q/2, so F′/F = (n − 1)²/(4n). Setting this equal to 16/9 gives 9n² − 82n + 9 = 0, whose solution greater than 1 is n = 9.

Question 15

A neutral metal sphere acquires a charge of +1.6 μC. Take the electron mass as 9.1 × 10⁻³¹ kg. As a result, the mass of the sphere
  1. increases by about 9.1 × 10⁻¹⁸ kg
  2. does not change at all
  3. decreases by about 9.1 × 10⁻¹⁸ kg
  4. decreases by about 1.7 × 10⁻¹⁴ kg
Show answer
Answer: (C) decreases by about 9.1 × 10⁻¹⁸ kg
Explanation: A metal becomes positively charged by losing electrons, not by gaining protons. The number lost is 1.6 × 10⁻⁶ / 1.6 × 10⁻¹⁹ = 10¹³. The mass lost is 10¹³ × 9.1 × 10⁻³¹ kg = 9.1 × 10⁻¹⁸ kg.

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