Question

Calculate the electrical force that acts on one plate of a parallel plate capacitor. The potential difference between the plates is 10 volts, and the plates are squares 20 cm on a side with a separation of 3 cm. If the plates are insulated so the charge cannot change, how much external work could be done by letting the plates come together

Answer 1

Answer:

The work done in bringing the plates together is 5.9 x 10⁻¹⁰ J.

Explanation:

Given;

potential difference between the plates, V = 10 V

length of each square side of the plates, L = 20 cm = 0.2 m

area of the plates, A = 0.2 x 0.2 = 0.04 m²

separation of the plates, d = 3 cm = 0.03 m

The work done in bringing the plates together is calculated as;

W = ¹/₂qV

$W = \frac{1}{2} (\frac{\epsilon_0 A }{d}V) \times V\\\\W = \frac{\epsilon_0 A V^2}{2d}\\\\W = \frac{8.85 \times 10^{-12} \ \times \ 0.04\ \times \ 10^2}{2(0.03)} \\\\W = 5.9 \times 10^{-10} \ J$

Therefore, the work done in bringing the plates together is 5.9 x 10⁻¹⁰ J.