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Bogdan [553]
2 years ago
6

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 magni

tude of 4 x10^-6 C the potential difference across the plates is approximately:
Physics
1 answer:
barxatty [35]2 years ago
5 0

Answer:

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;

V = \frac{Q}{C} = \frac{Qd}{\epsilon_o A} = \frac{(4*10^{-6})(0.1*10^{-3})}{(8.85*10^{-12})(0.2)}\\\\V = 226 \ V

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

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