Question

A parallel-plate capacitor has square plates that have edge length equal to 1.20×102cm and are separated by 1.00 mm. It is connected to a battery and is charged to 12.0 V. 1) How much energy is stored in the capacitor? (Express your answer to three significant figures.)

Answer 1

Answer:

Energy stored in the capacitor will be $1.835\times 10^{-6}J$

Explanation:

We have given edge length of capacitor = $1.2\times 10^2cm=1.2m$

So area A = 1.2×1.2 = 1.44 $m^2$

Separation is given as d  1 mm = $10^{-3}m$

We know that capacitance is given by $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 1.44}{10^{-3}}=12.744\times 10^{-9}F$

Voltage is given as V = 12 volt

We know that energy stored is given by $E=\frac{1}{2}CV^2=\frac{1}{2}\times 12.744\times 10^{-9}\times 12^2=1.835\times 10^{-6}J$

Answer 2

Answer:

9.18 x 10^-7 J

Explanation:

length of edge, s = 1.2 x 10^2 cm = 1.2 m

Separation between the plates, d = 1 mm = 0.001 m

V = 12 V

1. Formula for the capacitance of parallel plate capacitor is

$C=\frac{\epsilon _{0}A}{d}$

Where, A be the area of plates

A = side x side = 1.2 x 1.2 = 1.44 m^2

$C=\frac{8.854\times 10^{-12}\times 1.44}{0.001}$

C = 1.275 x 10^-8 F

Energy stored in a capacitor

$U = \frac{1}{2}CV^{2}$

$U = \frac{1}{2}\times 1.275 \times 10^{-8}\times 12^{2}$

U = 9.18 x 10^-7 J