Question

A dielectric slab of dielectric constant $K=12$ is inserted between two conducting plates having opposite charge densities of same magnitude on them. Find the magnitude of ratio of charge density induced on the surface of dielectric slab facing the conducting plate and the charge density on electric plate.

• A. $\frac{11}{12}$
• B. $\frac{1}{12}$
• C. $12$
• D. $\frac{12}{13}$

Answer 1

The correct option is A $\frac{11}{12}$

Let $E_0$ be the electric field without dielectric medium.

Then when dielectric medium is introduced, the electric field becomes , $E_{net}=\frac{E_0}{K}$

But, $E_{net}=E_0-E_{in}$ , where $E_{in}$ is induced electric field.

$\Rightarrow E_{in}=E_0-E_{net}$

$\Rightarrow E_{in}=E_0-\frac{E_0}{K}$

Here $E=\frac{\sigma }{{\epsilon }_0}$

$\Rightarrow {\sigma }_{in}=\left(\frac{K-1}{K}\right)\sigma$

$\Rightarrow \frac{{\sigma }_{in}}{\sigma }=\frac{11}{12}$

Hence, option $\left(a\right)$ is correct.