91影视

The capacitor is a charge and electrical energy storage device which consists of two parallel plates.

The capacitance of a capacitor is given by the following:

\(C = \frac{Q}{V} = {\varepsilon _0}\frac{A}{d}\) 鈥� (i)

Here, Q is the charge;V is the voltage;A is the area of plates;d is the separation between the plates.

The amount of electric potential energy stored inside the charged capacitor is given as follows:

\(PE = \frac{1}{2}QV = \frac{1}{2}C{V^2} = \frac{1}{2}\frac{{{Q^2}}}{C}\) 鈥� (ii)

The energy stored in a capacitor is given by the following:

\(PE = \frac{1}{2}C{V^2}\)

When the potential difference is doubled, the potential energy stored will be as follows:

\(\begin{array}{c}PE' = \frac{1}{2}C{{V'}^2}\\ = \frac{1}{2}C{\left( {2V} \right)^2}\\ = 4\left( {\frac{1}{2}C{V^2}} \right)\\ = 4PE\end{array}\)

Thus, the energy stored inside the capacitor becomes four times the initial value

When the separation of plates is doubled, i.e.,new separation is\(d' = 2d\) , charge Q on the plates will remain the same but capacitance gets changed.

Using equation (i),the new capacitance of the capacitor becomes the following:

\(\begin{array}{c}C' = {\varepsilon _0}\frac{A}{{d'}}\\ = {\varepsilon _0}\frac{A}{{\left( {2d} \right)}}\\ = \frac{1}{2}C\end{array}\)

Thus, using (ii), the amount of energy stored inside the capacitor becomes the following:

\(\begin{array}{c}PE' = \frac{1}{2}\frac{{{Q^2}}}{C}\\ = \frac{1}{2}\frac{{{Q^2}}}{{\left( {\frac{1}{2}C} \right)}}\\ = 2\left( {\frac{1}{2}\frac{{{Q^2}}}{C}} \right)\\ = 2PE\end{array}\)

Thus, the energy stored inside the capacitor doubles.