Question

A 12 pf capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Answer 1

Hint: To solve this question we need to obtain the mathematical expression for electrostatic energy stored in a capacitor in terms of the capacitance and the applied voltage. Put the values given in the question to obtain the required answer.

Complete step by step answer:
Given, the capacitance of the capacitor is 12 pF.
Now, one picofarad is equal to ${{10}^{-12}}$ farad.
So, capacitance can be written as,
$C=12pF=12\times {{10}^{-12}}F$
The capacitor is connected to a battery of voltage $V=12V$ .
The energy stored in a capacitor is electrostatic potential energy and it is related to the charge and voltage of the capacitor plates.
The electrostatic energy stored in a capacitor is given by the mathematical expression,
$E=\dfrac{1}{2}C{{V}^{2}}$
Where, C is the capacitance of the capacitor and V is the voltage applied on the capacitor.
Putting the given values on the above equation, we get that,
$\begin{aligned} & E=\dfrac{1}{2}\times 12\times {{10}^{-12}}\times 50 \\\ & E=3\times {{10}^{-9}}J \\\ \end{aligned}$
So, the energy stored in the capacitor is $3\times {{10}^{-9}}J$.

Additional information:
A capacitor consists of two plates. When a voltage is applied between the two plates it creates a potential difference between the two plates and an electric field is formed. This electric field moves the electron from the positive plate to the negative plate of the capacitor. We have positive charge on one side and negative charge on the other side of the capacitor. This electric field holds the potential energy in the capacitor.

Note:
The electrostatic energy stored in a capacitor can be expressed in many forms.
${{E}_{c}}=\dfrac{1}{2}{{V}^{2}}C=\dfrac{1}{2}\dfrac{{{Q}^{2}}}{C}=\dfrac{1}{2}QV$
Where, Q is the charge on the capacitor, V is the potential difference applied to the capacitor and C is the capacitance of the capacitor.