Question

A parallel-plate capacitor is formed from two 7.20 cm× 7.20 cm electrodes spaced 1.90 mm apart. The electric field strength inside the capacitor at the center is 3.00 ×10^6N/CWhat is the charge (in nC) on each electrode?

Answer 1

To solve this problem we will apply the concept related to the electric field as a function of the charge, the area and the permittivity constant of free space. This is mathematically defined as

$E= \frac{Q}{A\epsilon_0}$

Here,

Q= Charge

A = Area

$\epsilon_0$ = Permittivity of free space

Rearranging to find the charge,

$Q = A\epsilon_0 E$

Replacing,

$Q = (7.2*10^{-2})^2 (8.85*10^{-12})(3*10^6)$

$Q = 1.37635*10^{-7}C$

Therefore the charge is $1.37*10^{-7}C$