Two circular metal plates each of radius ' r ' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is ' C 1 '. If the radius of

Question

Two circular metal plates each of radius ' r ' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is ' C 1 ​ '. If the radius of each of the plates is increased to 2 ​ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ' C 2 ​ '. The ratio of the capacitances C 1 ​ : C 2 ​ . is

TestHub · Accepted Answer

The capacitance of a parallel plate capacitor is given by $C = \frac{ε _{0} A}{d} = \frac{ε _{0} π r ^{2}}{d}$ .

For the initial capacitor:
$C_{1} = \frac{ε _{0} π r ^{2}}{d}$

For the final capacitor, the radius is $r^{'} = 2 r$ and the distance is $d^{'} = \frac{d}{2}$ .

The new capacitance is:
$C_{2} = \frac{ε _{0} π ( 2 r ) ^{2}}{\frac{d}{2}} = \frac{ε _{0} π ( 2 r ^{2} )}{\frac{d}{2}} = \frac{4 ε _{0} π r ^{2}}{d}$

Taking the ratio of $C_{1}$ to $C_{2}$ :
$\frac{C _{1}}{C _{2}} = \frac{\frac{ε _{0} π r ^{2}}{d}}{\frac{4 ε _{0} π r ^{2}}{d}} = \frac{1}{4}$

The ratio $C_{1} : C_{2}$ is $1 : 4$ .

Answer: $1 : 4$

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