Question

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. 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. $6.5\, V$

Answer 1

Answer: (C)

The capacitor is charged by a battery of $25 V$. Let the magnitude of surface charge density on each plate be $\sigma$. Before inserting the dielectric slab, electric field strength between the plates,
$E =\frac{\sigma}{\varepsilon_{0}}=\frac{ V }{ d }$
or $\quad E=\frac{\sigma}{\varepsilon_{0}}=\frac{25}{5 \times 10^{-3}}$
$=5000 N / C$
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 $3 mm$ is placed between the plates, the thickness of air between the plates will be $5-3=2 mm .$ Electric field strength in air will have the same value $(5000 N / C )$ but inside the dielectric, it will be $\frac{5000}{ K }=\frac{5000}{10}=500 N / C$
so potential difference $= E _{\text {air }} d _{\text {air }}+ E _{\text {med }} d _{\text {med }}$
$=\left(5000 \times 2 \times 10^{-3}\right)+\left(500 \times 3 \times 10^{-3}\right)$
$=11.5\, V$