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Electric Charges and Fields · Question 20

JEE Main 2025 (Online) 2nd April Morning ShiftMCQEasy📈 Topic in 10% papers+4−1
1

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density + and -2 . The force experienced by a point charge +q placed at the mid point between two plates will be: [figure]

  1. 3 q4 0
  2. 3 q2 0
  3. q4 0
  4. q2 0

Answer

(2) 3 q2 0

Solution

Field of a large plane sheet of charge density σ: E = 2 0 (independent of distance).

Between the plates, the field of + points away from it and the field of -2 points towards it, i.e. both point the same way (towards the -2 plate):

E = 2 0 + 2 2 0 = 3 2 0

(Same result with charge redistribution on the conductors: inner faces carry + 3 2 and - 3 2, giving E = 3 2 0.)

F = qE = 3 q2 0, directed towards the -2 plate.

Answer: (2)

JEE Main 2025 (Online) 2nd April Morning Shift

Keep practising — Electric Charges and Fields

2

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

  1. (3d/4, 0, 0)
  2. (d/4, 0, 0)
  3. (4d/3, 0, 0)
  4. (d/3, 0, 0)
JEE Main 2025 (Online) 2nd April Morning Shift
3

A small bob of mass 100 mg and charge +10 is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density ' ' as shown in figure. If string subtends an angle of 45 with the sheet at equilibrium the charge density of sheet will be: (Given, 0 = 8.85 10-12 Fm and acceleration due to gravity, g = 10 ms2) [figure]

  1. 1.77 nC/m2
  2. 0.885 nC/m2
  3. 885 nC/m2
  4. 17.7 nC/m2
JEE Main 2025 (Online) 2nd April Morning Shift
4

Consider a circular loop that is uniformly charged and has a radius a 2. Find the position along the positive z-axis of the Cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in xy plane at the origin :

  1. a
  2. a/2
  3. 0
  4. a/ 2
JEE Main 2025 (Online) 2nd April Evening Shift
5

An electric dipole of mass m, charge q, and length l is placed in a uniform electric field E = E0 i. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :

  1. 12 ml2qE0
  2. 2 ml2qE0
  3. 2 mlqE0
  4. 12 2mlqE0
JEE Main 2025 (Online) 29th January Morning Shift
6

Match List - I with List - II. Choose the correct answer from the options given below :

  1. (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
  2. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  3. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  4. (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
JEE Main 2025 (Online) 29th January Morning Shift

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