Answer to Question #149477 in Physics for Nwose Sammy

Question

The plates if parallel plate capacitor are 5mm apart and 2m² in area. The plates are in vacuum. A potential difference of 1kv is applied across the capacitor. Calculate a. The capacitance.b. charge on each plate .c. Electric intensity in the space between them. d. Energy stored in the capacitor.

Answer 1

a)

$$C=\frac{\epsilon_0A}{d}=\frac{8.85\cdot10^{-12}\cdot2}{5\cdot10^{-3}}=3.54\cdot10^{-9}F$$

b)

$$q=CV=1000\cdot 3.54\cdot10^{-9}=3.54\cdot10^{-6}C$$

c)

$$E=\frac{\sigma}{\epsilon_0}=\frac{q}{\epsilon_0 A}\\ E=\frac{3.54\cdot10^{-6}}{2\cdot 8.85\cdot10^{-12}}=2\cdot10^{5}\frac{N}{C}$$

d)

$$W=0.5CV^2=0.5(1000)^2\cdot 3.54\cdot10^{-9}\\W=0.00177\ J=1.77\ mJ$$