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Electrostatic Potential and Capacitance

JEE Main 2025 (Online) 23rd January Evening ShiftMCQEasy📈 Topic in 5% papers+4−1
Ques. 1JEE Main 2025, 23 January (Evening Shift)

Two point charges -4 and 4 , constituting an electric dipole, are placed at (-9,0,0) cm and (9,0,0) cm in a uniform electric field of strength 104 N C-1. The work done on the dipole in rotating it from the equilibrium through 180 is :

  1. 18.4 mJ
  2. 12.4 mJ
  3. 16.4 mJ
  4. 14.4 mJ

Answer

(4) 14.4 mJ

Solution

Dipole moment p=q d=4 10-6 0.18=7.2 10-7 C m (separation 18 cm).

[figure]

Turning from stable equilibrium ( =0 , U=-pE) to =180 (U=+pE): W=2pE

W=2 7.2 10-7 104=14.4 mJ

Answer: (4)

Keep practising — Electrostatic Potential and Capacitance

Ques. 2JEE Main 2025, 23 January (Evening Shift)

Two charges 7 and -4 are placed at (-7 cm,0,0) and (7 cm,0,0) respectively. Given, 0=8.85 10-12 C2 N-1 m-2, the electrostatic potential energy of the charge configuration is :

  1. -2.0 J
  2. -1.5 J
  3. -1.2 J
  4. -1.8 J
Ques. 3JEE Main 2025, 22 January (Morning Shift)

A parallel-plate capacitor of capacitance 40 is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K=2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

  1. 8 mC and 2.0 J
  2. 4 mC and 0.2 J
  3. 2 mC and 0.2 J
  4. 2 mC and 0.4 J
Ques. 4JEE Main 2025, 22 January (Evening Shift)

For a short dipole placed at origin O , the dipole moment P is along x-axis, as shown in the figure. If the electric potential and electric field at A are V0 and E0, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y-axis is given by [figure]

  1. V02 and E016
  2. zero and E08
  3. V0 and E04
  4. zero and E016

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