A parallel-plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm.If the charge on each plate has a magnitude of 4 *10^-6 C the potential difference across the plates is approximately: A)0 V B)4 *10^-2 V C)2 *10² V D)2 * 10⁵ V E)4 *10⁸ V

Question

Quizplus · Accepted Answer

The potential difference across the plates of a capacitor is given by:

V = Q/C

where Q is the charge on each plate and C is the capacitance of the capacitor.

Here, Q = 4×10^(-6) C and C = εA/d, where ε is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.

Substituting the given values, we get:

C = εA/d = (8.85×10^(-12) F/m)(0.2×10^(-3) m^2)/(0.1×10^(-3) m) = 1.77×10^(-11) F

Therefore, V = Q/C = (4×10^(-6) C)/(1.77×10^(-11) F) ≈ 2.26×10^2 V ≈ 2.0×10^2 V

So, the best answer choice is C, which is approximately 2.0×10^2 V.

A Parallel-Plate Capacitor Has a Plate Area of 0