### JEE Main On 08 April 2018 Question 12

##### Question: A parallel plate capacitor of capacitance $ \text{90 pF} $ is connected to a battery of $ \text{emf 20 V} $ . If a dielectric material of dielectric constant $ \text{K=}\frac{5}{3} $ is inserted between the plates, the magnitude of the induced charge will be: [JEE Main Online 08-04-2018]

#### Options:

A) $ 2\text{.4 n C} $

B) $ 0\text{.9 n C} $

C) $ 1\text{.2 n C} $

D) $ 0\text{.3 n C} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

$ q=CV=3\times {10^{-9}};hereC=KC _0 $

$ q _{ind}=3\times {10^{-9}}( 1-\frac{1}{K} ) $

$ q _{ind}=3\times {10^{-9}}( 1-\frac{3}{5} )=3\times {10^{-9}}\times \frac{2}{5} $

$ q _{ind}=1.2\times {10^{-9}} $