Question

1 : - The distance between the two plates of a capacitor is doubled and the area of each plate is halved what happens to the capacitance.​

Answer 1

Given :

Distance between two plates is doubled and area of each plate is halved$.$

To Find :

Change in capacitance of capacitor$.$

SoluTion :

Let initial capacitance of capacitor be C.

Formula of capacitance in terms of distance between two plates (d) and area of each plate (A) is given by

• C = ε°A / d

Here, Permittivity of free space is constant so we can say that, Capacitance of capacitor is directly proportional to the distance b/w two plates and inversely proportional to the area of each plate.

$\leadsto\tt\:\dfrac{C}{C'}=\dfrac{A}{d}\times \dfrac{d'}{A'}$

$\leadsto\tt\:\dfrac{C}{C'}=\dfrac{A}{d}\times \dfrac{2d\times 2}{A}$

$\leadsto\tt\:\dfrac{C}{C'}=4$

$\leadsto\underline{\boxed{\mathbb{\purple{C'=\dfrac{C}{4}}}}}$

Answer 2

We know that , the capacitance of parallel palte capacitor is given by

$\boxed{ \sf{C = \frac{ e_{o}A}{d} }}$

Where ,

A = Area of each capacitor

d = Distance between two plates

Now , If distance between the two plates of a capacitor is doubled and the area of each plate is halved , then the new capacitance will be

$C' = \frac{ e_{o}}{2d} \times \frac{A}{2} \\ \\ C' = \frac{ e_{o}A}{4d} \\ \\ C' = \frac{C}{4}$

Therefore ,

• The new capacitance is 1/4 times of the initial capacitance