Question

A dielectric of dielectric constant 5.0 filled in the gap between the plates of a capacitor. Calculate the factor by which thae capacitance is increased, if the dielectric is only sufficient to fill up 1/5 of the gap.

Answer 1

As capacitance with dielectric in between the plates is given by:

$C’= \cfrac{A \epsilon_o}{d-t+\cfrac{t}{K}}$

here , d is separation between plates, and t is the thickness of dielectric .

So on putting the values we get:

$C’= \cfrac{A \epsilon_o}{d-\cfrac{d}{5}+\cfrac{d}{5\times5 }}$

$C’= \cfrac{25A \epsilon_o}{21d}$

And capacitance without dielectric is given by :

$C= \cfrac{A \epsilon_o}{d}$

So the ratio of capacitance would be :

$\cfrac{C’}{C}= \cfrac{25}{21}$

hence by a factor of $\cfrac{25}{21}$ The capacitance increase .