Answer to Question #214146 in Electric Circuits for nela

Question

The plates of parallel-plate capacitor in air are 6.0 mm apart and 2.5 cm (12) 2 in area. A potential difference of 5 kV is applied across the capacitor. Find:

(i) the capacitance of the capacitor,

(ii) the magnitude of charge on each plate,

(iii) the magnitude of the electric field between the plates, and

(iv) the energy stored in the capacitor.

Answer 1

Parralled plate capacitor

b=6.0mm

d=2.5cm

$\epsilon=12$

Area(A)=2cm²

V=5kV

Part (1)

$C=\frac{A\epsilon_0}{d-\frac{t}\epsilon_r}$

$$C=\frac{2\times8.85\times10^{-12}}{2.5\times10^{-2}-\frac{6\times10^{-3}}{12}}=7.22\times10^{-10}F$$

$C=0.722nF$

Part(b)

Q=CV

$$Q={7.22\times10^{-10}\times 5\times 10^{3}}=3.61\times10^{-6}$$

Q=$3.61\mu c$

Part(c)

$$E=\frac{V}{l}=\frac{5\times10^3}{2.5\times10^{-2}}=20V/m$$

Part(d)

$U=\frac{1}{2}QV$

$$U=\frac{1}{2}\times3.61\times10^{-6}\times5\times10^3=9.025mJ$$