StreakPeaked· Practice

ExamsJEE AdvancedPhysics

Three large thick conducting parallel plates have charges 20 microC, -5 microC, and -15 microC respectively. The separation between plate 1 and plate 2 is 1 cm, and between plate 2 and plate 3 is 2 cm. Point A is in the region between plates 1 and 2, and point B is in the region between plates 2 and 3. Find the ratio of the electric field magnitude at B to that at A, i.e., E_B / E_A.

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 1

Solution

For a system of large conducting plates, the electric field in each gap depends only on the net charge enclosed between relevant surfaces. The charge on the inner faces of the outermost plates are equal and opposite (each = total charge / 2 signed). Total charge = 20 - 5 - 15 = 0. When total charge is zero, the outer face charges are both zero. Inner face charges are determined by the constraint that field inside conductor = 0. The field in region A (between plate 1 and plate 2): E_A = sigma_inner/(epsilon₀) where sigma_inner is from adjacent faces. By symmetry and charge distribution analysis, E_A = E_B, so E_B/E_A = 1.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →