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 $B$. Charge supplied by the battery is :

(Given ${\epsilon }_0=8.85\times {10}^{-12}N{m}^2{C}^{-2}$)

• A. $120pC$
• B. $100pC$
• C. $60pC$
• D. $50pC$

Answer 1

Answer: (B)

$100pC$

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

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: $\Delta q=(50+50)pC=100pC$