Question

A parallel plate capacitor has a capacity $80\times {10}^{-6}F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is :

• A. $45.6\times {10}^{-3}C$
• B. $25.3\times {10}^{-3}C$
• C. $120\times {10}^{-3}C$
• D. $12\times {10}^{-3}C$

Answer 1

The correct option is A $45.6\times {10}^{-3}C$

Charge passing through the wire, $\Delta q=\Delta CV$

$=(C^{\prime }-C)V=(k-1)CV=(20-1)(80\times {10}^{-6})\left(30\right)=45.6\times {10}^{-3}$

So, charge passing through the wire $=45.6\times {10}^{-3}C$