A parallel-plate capacitor has a plate area of 0.2 m^2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 4 x10^-6 C the potential difference across the plates is approximately:

The potential difference across the plates is 226 V.

Explanation:

Given;

area of the capacitor plate, A = 0.2 m²

separation, d = 0.1 mm = 0.1 x 10⁻³ m

charge on each plate, Q = 4 x 10⁻⁶ C

Charge on the capacitor is given by;

Q = CV

Where;

C is the capacitance of the capacitor, given as;

C = ε₀A / d

Then, the potential difference across the plates is given by;

Therefore, the potential difference across the plates is 226 V.

we need some answer choices, or something like that

explanation: