Question

The separation between the plates of a parallel plate capacitor is $5mm$ . This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference $25V$ using a battery. The battery is then disconnected and a dielectric slab of thickness $3mm$ and dielectric constant $K=10$ is placed between the plates, as shown. The potential difference between the plates after the dielectric slab has been introduced is

• A. $18.5V$
• B. $13.5V$
• C. $11.5V$
• D. $6.5V$

Answer 1

Answer: (C)

The capacitor is charged by a battery $25V$ . Let the magnitude of surface charge density on each plate be $\sigma$ . Before inserting the dielectric slab, field strength between the plates,
$E=\frac{\sigma }{{ϵ}_0}=\frac{V}{d}$
or $E=\frac{\sigma }{{ϵ}_0}=\frac{25}{5\times 10^{-3}}=5000N/C$
The capacitor is disconnected from the battery but the charge on it will not change so $\sigma$ has the same value. When a dielectric slab of thickness $3mm$ is placed between the plates, the air thickness between the plates will be $5-3=2mm$ . Electric field strength in air will have the same value $\left(5000N/C\right)$ but inside the dielectric, it will be $\frac{5000}{K}=\frac{5000}{10}=500N/C$
So potential difference
$V=E_{air}{d}_{air}+E_{med}{d}_{med}$
$V=5000\times \left(2\times 10^{-3}\right)+500\times \left(3\times 10^{-3}\right)$
$V=11.5V$