Question

Two 1.60cm×1.60cm plates that form a parallel-plate capacitor are charged to ± 0.711nC . What is the potential difference across the capacitor if the spacing between the plates is 1.20mm What is the potential difference across the capacitor if the spacing between the plates is 2.40mm ?

Answer 1

The surface charge density:
S(igma) = Q / A
A = 1.6 * 10^(-2) * 1.6 * 10^(-2) = 2.56 * 10^(-4) m²
The electric field:  E = S / E(psilon) o 
E = 0.711 * 10^^(-9) / ( 2.56 * 10 ^(-4) * 8.85 * 10 ^(-12) ) = 
= 0.3138 * 10 ^6 = 3.138 * 10^5 V/m
a ) Δ V = E d
Δ V = 3.138 * 10^5 V/m* 1.2 * 10^(-3) m = 376.56 V
b ) When d = 2.40 mm:
Δ V = 3.138 * 10^5 V/m * 2.40 *10^(-3) m = 753.12 V

Answer 2

Hope you are able to see the solution clearly :)