Question

A dielectric material of relative dielectric constant 6 is now inserted and completely fills the space between the plate capacitor.Calculate the new chargee

Answer 1

Let the capacitance of the capacitor be C.

After inserting the dielectric material in the capacitor, $(C_1)=kC$

Here, in the question it is given , $k=6$

$(C_1)=6C$

Total energy always be conserve,

So, $\frac{Q_1^2}{2C}=\frac{Q_2^2}{2C_2}$

Now, substituting the values,

$Q_2=\frac{Q_1^2 C_2}{C_1}$

$\Rightarrow Q_2=\frac{Q_1\sqrt{\times 6c}}{\sqrt{c}}$

Total charge always be conserve, so it will not change.