Question

The separation between the plates of a parallel plate capacitor is $5 \, mm$ . This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference $25 \, V$ using a battery. The battery is then disconnected and a dielectric slab of thickness $3 \, mm$ 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. $18.5 \, V$
• B. $13.5 \, V$
• C. $11.5 \, V$
• 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,
$E=\frac{\sigma }{\epsilon _{0}}=\frac{V}{d}$
or $E=\frac{\sigma }{\epsilon _{0}}=\frac{25}{5 \times 1 0^{- 3}}=5000N/C$
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(5000 N / C\right) \, $ but inside the dielectric, it will be $\frac{5000}{K}=\frac{5000}{10}=500N/C$
So potential difference
$V=E_{a i r}d_{a i r}+E_{m e d}d_{m e d}$
$V=5000\times \left(2 \times 1 0^{- 3}\right)+500\times \left(3 \times 1 0^{- 3}\right)$
$V=11.5 \, V$