Electric Field: NEET MCQs with Answers

15 free questions on Electric Field 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 field: Electric field due to a point charge, Electric field lines.

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Question 1

Previous-year question
An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of the electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton, is [NEET 2018]
  1. smaller
  2. 5 times greater
  3. equal
  4. 10 times greater
Show answer
Answer: (A) smaller
Explanation: Both are accelerated by the same force eE, but the electron's mass is about 1836 times smaller, so its acceleration is far larger and h = (1/2)at² gives a much shorter time of fall.

Question 2

Five identical charges of +2 nC are placed at the corners A, B, C, D and E of a regular pentagon. Each corner is 30 cm from the centre O of the pentagon. What is the electric field at O, first with all five charges present, and then after the charge at A alone is removed? (Take 1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. Zero at first, and then 200 N C⁻¹ directed from O towards A
  2. Zero at first, and then 200 N C⁻¹ directed from A towards O
  3. 1000 N C⁻¹ at first, and then 800 N C⁻¹ directed from O towards A
  4. Zero both times
Show answer
Answer: (A) Zero at first, and then 200 N C⁻¹ directed from O towards A
Explanation: Five equal fields directed along symmetric lines cancel at the centre of a regular polygon, so the field is zero at first. Removing the charge at A is the same as adding −2 nC at A to the complete set. That single negative charge gives E = 9 × 10⁹ × 2 × 10⁻⁹/(0.30)² = 200 N C⁻¹ at O, directed towards it, that is, towards A.

Question 3

Electric field lines leave a charged body and fan out into the surrounding space. Close to point A the lines are crowded together, while close to point B, a distance r away, they are widely spaced. If the field magnitudes at A and B are E_A and E_B, then
  1. E_A > E_B
  2. E_A = E_B/r
  3. E_A < E_B
  4. E_A = E_B
Show answer
Answer: (A) E_A > E_B
Explanation: The number of field lines crossing unit area held perpendicular to them measures the strength of the field. The lines are more crowded at A than at B, so the field is stronger at A. No simple formula in r connects the two values, because the body is not stated to be a point charge.

Question 4

A particle of mass 1 mg and charge +5 μC is released from rest at x = 0 in a region where the electric field acts along the x-axis and is given by E = (100 − 20x) N/C, with x in metres. Only this field acts on the particle. What is the greatest speed reached, and at what position?
  1. 50 m/s at x = 10 m
  2. 50 m/s at x = 5 m
  3. 25 m/s at x = 5 m
  4. 100 m/s at x = 5 m
Show answer
Answer: (B) 50 m/s at x = 5 m
Explanation: The force is qE, so the speed rises until E = 0, which happens at x = 5 m. The work done up to that point is q∫(100 − 20x)dx = q(100x − 10x²) = (5 × 10⁻⁶)(250) = 1.25 × 10⁻³ J. Setting this equal to ½mv² with m = 1 × 10⁻⁶ kg gives v² = 2500, so v = 50 m/s. Beyond x = 5 m the field reverses and the particle slows, stopping at x = 10 m.

Question 5

A small positive charge is released from rest at a point where the electric field lines of a fixed charge system are curved. Only the electric force acts on it. Which statement about its motion is correct?
  1. It starts along the tangent to the field line, but then does not stay on that line
  2. It stays at rest, because a curved field line exerts no net force on a charge
  3. Its acceleration stays perpendicular to the field line at every instant
  4. It moves exactly along the field line, since the force on it is always tangent to that line
Show answer
Answer: (A) It starts along the tangent to the field line, but then does not stay on that line
Explanation: The tangent to a field line gives the direction of the force, and so of the acceleration, not of the velocity. Starting from rest, the charge first moves along the tangent. Once it has velocity, the changing force cannot turn it sharply enough to keep it on a curved line, so its path departs from the line. A field line is the trajectory of a charge released from rest only when the line is straight.

Question 6

Two oil drops of the same density are each held stationary in air by the same vertical electric field. When the field is switched off, they fall with terminal speeds v and 4v. Ignoring the buoyancy of air, the ratio of the charge on the faster drop to that on the slower drop is
  1. 2
  2. 4
  3. 16
  4. 8
Show answer
Answer: (D) 8
Explanation: By Stokes' law, the terminal speed is proportional to r², so the faster drop has twice the radius. When balanced, qE = mg, and the mass is proportional to r³, so q is proportional to r³ for the same field. The charge ratio is therefore 2³ = 8.

Question 7

Point charges lie on the positive x-axis as follows: +1 μC at x = 1 m, −1 μC at x = 3 m, +1 μC at x = 9 m, −1 μC at x = 27 m, and so on without end. The electric field at the origin is (take 1/(4πε₀) = 9 × 10⁹ N m² C⁻²)
  1. 10125 N/C along the −x direction
  2. 8100 N/C along the −x direction
  3. 9000 N/C along the −x direction
  4. 8100 N/C along the +x direction
Show answer
Answer: (B) 8100 N/C along the −x direction
Explanation: The nearest charge alone gives kq/r² = 9000 N/C, directed away from it, along −x. Each later charge alternates in sign and contributes one-ninth of the previous term, so E = 9000(1 − 1/9 + 1/81 − ...) along −x. This geometric series sums to 1/(1 + 1/9) = 9/10, so E = 8100 N/C.
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Question 8

Two particles of equal mass, with charges +3q and −q, are free to move in a uniform electric field E in vacuum, with the +3q particle ahead of the −q particle along the direction of E. Gravity is negligible. If they accelerate together while keeping a constant separation L, then L equals
  1. √(3q / (8πε₀E))
  2. √(3q / (16πε₀E))
  3. √(3q / (4πε₀E))
  4. √(q / (4πε₀E))
Show answer
Answer: (A) √(3q / (8πε₀E))
Explanation: The net external force is 3qE − qE = 2qE on a total mass 2m, so a = qE/m along the field. For the −q particle, F − qE = ma = qE, where F is the mutual attraction, so F = 2qE. Setting 3q²/(4πε₀L²) = 2qE gives L² = 3q/(8πε₀E).

Question 9

A proton and an α-particle enter the same uniform transverse electric field of fixed length, each moving perpendicular to the field. What is the ratio of the sideways deflection of the proton to that of the α-particle, first when they enter with the same speed and then when they enter with the same kinetic energy?
  1. 2 : 1 at the same speed, and 1 : 1 at the same kinetic energy
  2. 1 : 2 at the same speed, and 2 : 1 at the same kinetic energy
  3. 1 : 1 at the same speed, and 1 : 2 at the same kinetic energy
  4. 2 : 1 at the same speed, and 1 : 2 at the same kinetic energy
Show answer
Answer: (D) 2 : 1 at the same speed, and 1 : 2 at the same kinetic energy
Explanation: With field length L, the transit time is L/v and the sideways acceleration is qE/m. The deflection is y = qEL²/(2mv²). At equal speed y is proportional to q/m, which is e/m for the proton and e/2m for the α-particle, giving 2 : 1. Writing mv² = 2K gives y = qEL²/(4K), so at equal kinetic energy y is proportional to q, giving 1 : 2.

Question 10

Charges +q are fixed at five vertices of a regular hexagon of side a, and a charge −2q is fixed at the sixth vertex. The electric field at the centre of the hexagon has magnitude
  1. 3q/(4πε₀a²), directed towards the −2q charge
  2. q/(4πε₀a²), directed towards the −2q charge
  3. 3q/(4πε₀a²), directed away from the −2q charge
  4. 2q/(4πε₀a²), directed away from the −2q charge
Show answer
Answer: (A) 3q/(4πε₀a²), directed towards the −2q charge
Explanation: Treat the −2q charge as +q together with −3q at the same vertex. Six equal charges +q at the vertices give zero field at the centre by symmetry. Only −3q remains, at a distance a from the centre because the vertex distance of a regular hexagon equals its side. Its field has magnitude 3q/(4πε₀a²) and points towards the negative charge.

Question 11

A proton, a deuteron and an alpha particle are each accelerated from rest through the same potential difference. Each then enters a uniform electric field at right angles to the field, and travels the same distance along its original direction. How do their sideways deflections compare?
  1. The deuteron deflects the most
  2. The alpha particle deflects half as much as the proton
  3. All three deflect by the same amount
  4. The alpha particle deflects twice as much as the proton
Show answer
Answer: (C) All three deflect by the same amount
Explanation: The sideways deflection after a distance L is y = qEL²/(2mu²) = qEL²/(4K), where K is the kinetic energy. Acceleration through a potential difference V gives K = qV. Substituting gives y = EL²/(4V), which contains neither the charge nor the mass. All three particles therefore follow the same path.

Question 12

Point charges of magnitude 10 nC are fixed on the x-axis at x = 1 m, 3 m, 9 m, 27 m and so on without end. The signs alternate, and the charge at x = 1 m is positive. What is the electric field at the origin? (1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. 81 N/C along −x
  2. 81 N/C along +x
  3. 90 N/C along −x
  4. 101.25 N/C along −x
Show answer
Answer: (A) 81 N/C along −x
Explanation: A positive charge on the +x side gives a field along −x at the origin, and a negative one gives a field along +x. Taking −x as positive, E = kq(1 − 1/9 + 1/81 − ...) with distances in metres. The geometric series sums to 1/(1 + 1/9) = 9/10, and kq = 90 N/C, so E = 81 N/C along −x.

Question 13

A small charged bead starts from rest on a smooth horizontal insulating rod, with a uniform electric field along the rod. It reaches a speed of 8 m/s in 2 s. At that instant the field is reversed, keeping its magnitude, and it stays reversed. At about what time, measured from the start, does the bead pass its starting point again?
  1. 6.8 s
  2. 6.0 s
  3. 4.0 s
  4. 8.0 s
Show answer
Answer: (A) 6.8 s
Explanation: The acceleration has magnitude 4 m/s² throughout. The bead covers 8 m in the first 2 s, then slows to rest in another 2 s after a further 8 m, so at t = 4 s it is 16 m from the start. Returning from rest, 16 = ½ × 4 × τ² gives τ = 2√2 ≈ 2.8 s. The total time is about 6.8 s.

Question 14

A proton and an alpha particle enter, in turn, the uniform electric field between two parallel charged plates, each moving parallel to the plates. Gravity is negligible and each stays between the plates along their full length. What is the ratio of the sideways deflection of the alpha particle to that of the proton (i) if they enter with the same kinetic energy and (ii) if they enter with the same speed?
  1. (i) 1 : 1; (ii) 1 : 2
  2. (i) 2 : 1; (ii) 1 : 2
  3. (i) 2 : 1; (ii) 2 : 1
  4. (i) 1 : 2; (ii) 2 : 1
Show answer
Answer: (B) (i) 2 : 1; (ii) 1 : 2
Explanation: The particle crosses plate length L in time L/v with sideways acceleration qE/m, so the deflection is y = qEL²/(2mv²). Written with kinetic energy K, this is y = qEL²/(4K), so at equal K the deflection goes as q, giving 2 : 1. At equal speed the deflection goes as q/m, which for the alpha particle is 2e/4m against e/m for the proton, giving 1 : 2.

Question 15

Two point charges exert an electrostatic force of 0.9 N on each other. The electric field at the position of the first charge, due to the second, is 9 × 10⁴ N/C, and the field at the position of the second charge, due to the first, is 2.25 × 10⁴ N/C. What is the separation between the charges? (1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. 1 m
  2. 2 m
  3. 0.5 m
  4. 4 m
Show answer
Answer: (B) 2 m
Explanation: The force on each charge is its charge times the field at its position, and the two forces are equal by Newton's third law. So q₁ = 0.9/(9 × 10⁴) = 1 × 10⁻⁵ C and q₂ = 0.9/(2.25 × 10⁴) = 4 × 10⁻⁵ C. Then d² = kq₁q₂/F = 9 × 10⁹ × 4 × 10⁻¹⁰ / 0.9 = 4 m², so d = 2 m.

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