Question

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