Question

the capacitance of a parallel plate capacitor is 2.5uf when it is half filled with a dielectric as shownnin fig, its capacitance becomes 5uf, the dielectric constant of dielectric is​

Answer 1

Given:

The capacitance of a parallel plate capacitor is 2.5uF. When it is half filled with a dielectric, its capacitance becomes 5uF.

To find:

Value of dielectric constant .

Calculation:

Initially , the capacitance without dielectric slab :

$\therefore \: 2.5 = \dfrac{A\epsilon_{0}}{d}$

Finally , capacitance with dielectric slab :

$\therefore \: 5 = \dfrac{ \frac{A}{2}\epsilon_{0}}{d} + k \bigg(\dfrac{ \frac{A}{2}\epsilon_{0}}{d} \bigg)$

$= > \: 5 = \dfrac{1}{2} \bigg( \dfrac{ A\epsilon_{0}}{d} \bigg) + \dfrac{k}{2} \bigg(\dfrac{ A\epsilon_{0}}{d} \bigg)$

$= > \: 5 = \bigg \{ \dfrac{1}{2} + \dfrac{k}{2} \bigg \} \bigg(\dfrac{ A\epsilon_{0}}{d} \bigg)$

$= > \: 5 = \bigg \{ \dfrac{1}{2} + \dfrac{k}{2} \bigg \} \times 2.5$

$= > \: \bigg \{ \dfrac{1}{2} + \dfrac{k}{2} \bigg \} = 2$

$= > \: \dfrac{k}{2} = \dfrac{3}{2}$

$= > k = 3$

So , final answer is :

Dielectric constant of slab is 3.