Question

A parallel plate capacitor has rectangular plates of 400 cm^2 area and are separated by a distance of 2 mm with air as the medium. What charge will appear on the plates if a 200-volt potential difference is applied across the capacitor?

Answer 1

Step 1: Given

Area of plate(A)= 400 cm² = 0.04 m²

Distance(d)= 2 mm= 0.002 m

Potential difference(V)= 200 V

$\epsilon$ is constant and Q are charges.

Step 2: Formula used

Capacitance(c)= $\frac{\epsilon A}{d}$

Charge(Q)= cV

Step 3: Finding the charges that appear on the plate

Q=$\frac{\epsilon A}{d}$*V

Q= $\frac{8.854\times {10}^{-12}\times 0.04\times 200}{0.002}$ C

Q= $3.54\times {10}^{-8}C$

Hence, the charges that appear on the plate are $3.54\times {10}^{-8}C$.