Question

Each plate of a parallel capacitor has area S = $5\times {10}^{-3}{m}^2$ and the distance between the plates is d = 8.80 mm apart. Plate A has a positive charge $q_1={10}^{–10}$ C and plate B has charge $q_2=+2\times {10}^{–10}$ C. Energy supplied by a battery of emf E = 10 volt when its positive terminal is connected with plate A and negative terminal with plate B is $n\times {10}^{–9}$ Joule. Find n. Given $({\epsilon }_0=8.85\times {10}^{–12}{N}^{-1}{m}^{-2}{C}^2)$

Answer 1

We know that –
C = $\frac{{\epsilon }_{0}A}{d}$
Here, S = A = $5\times {10}^{-3}{m}^2$
$q_1={10}^{-10}C,q_2=2\times {10}^{-10}$ C
E = 10 V, Energy supplied = ?
C = $\frac{8.8\times {10}^{-12}\times 5\times {10}^{-3}}{8.8\times {10}^{-3}}=5\times {10}^{-12}F$
Charge on the plate after connection with the battery is

q = CV = $5\times {10}^{-12}\times 10=50$ pC
Charge supplied by the battery is
$\Delta$ q = (50 + 50) pC = 100 pC
Energy supplied by the batttery is
$U_{battery}=\Delta$ QV = 100 $\times$ 10
= 1000 $\times$ pJ = 1 $\times {10}^{-9}$ J.