Question

Each plate of a parallel plate air capacitor has area $S=5\times {10}^{-3}{m}^2$ and the distance between the plates is $d=8.80mm$. Plate $A$ has positive charge $q_1=+{10}^{-10}C$, and plate $B$ has charge $q_2=+2\times {10}^{-10}C$. A battery of emf $E=10V$ has its positive terminal connected to plate $A$ and the negative terminal to plate $B$. Energy supplied by the battery is :

(Given ${\epsilon }_0=8.8\times 1^{-12}N{m}^2{C}^{-2}$)

• A. ${10}^{-9}J$
• B. $5\times {10}^{-9}J$
• C. $50\times {10}^{-9}J$
• D. $25\times {10}^{-9}J$

Answer 1

The correct option is A ${10}^{-9}J$

$C=\frac{{\epsilon }_{0}A}{d}=\frac{8.8\times {10}^{-12}\times 5\times {10}^{-3}}{8.8\times {10}^{-3}}=5\times {10}^{-12}F$

Charge on the plate after connection with the battery is: $q=CV=5\times {10}^{-12}\times 10=50pC$

Charge supplied by the battery is: $\Delta q=(50+50)pC=100pC$

Energy supplied by the battery is: $U_{battery}=\Delta qV=100\times 10=1000pJ$ $={10}^{-9}J$