Question

Two plates A and B of a parallel plate capacitor are arranged in such a way, that the area of each plate is S = 5 × 10^-3 m 2 and distance between them is d = 8.85 mm. Plate A has a positive charge q₁ = 10^-10 C and Plate B has charge q₂ = + 2 × 10^-10 C. Then the charge induced on the plate B due to the plate A be - (....... × 10^-11 )C

Options

• 5

• 6

• 7

• 8

Answer 1

5

Explanation:

10-¹⁰ - x x - x 2 × 10^-10 + x

Let the induced charge be x,

At the steady state, the potential will be equal, we know that potential due to charged plate V = `["Q"/2(epsilonxx"S")]xx"d"`

10^-10 - x = 2 × 10^–10 + x

2x = - 10^-10

x = - 5 × 10^–11 C