Question

The capacity of a parallel plate condenser is $5\mu F$.When a glass plate is placed between the plates of the condenser,its potential difference reduces to 1/8 of original value.The magnitude of relative dielectric constant of glass is

• A. 2
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is D 8

Potential of a capacitor without dielectric, $V=\frac{Q}{C}=\frac{Qd}{A{\epsilon }_0}$ where

$Q$ is the charge on the capacitor

$d$ is the plate distance

$A$ is the area of plates

${\epsilon }_0$ is permeability of free space

After insertion of dielectric,potential becomes:

$V^{\prime }=\frac{Q}{C^{\prime }}$=$\frac{Qd}{kA{\epsilon }_0}$ where k is dielectric constant.

since, $V^{\prime }=\frac{V}{8}$

$\frac{Qd}{kA{\epsilon }_0}=\frac{Qd}{8A{\epsilon }_0}$

$k=8$