Question

A 16pF capacitor is connected to \[70V\] supply. The amount of electric energy stored in the capacitor is:-
A. \[4.5 \times {10^{ - 12}}J\]
B. \[5.6 \times {10^{ - 8}}J\]
C. \[2.5 \times {10^{ - 12}}J\]
D. \[3.9 \times {10^{ - 8}}J\]

Answer 1

Hint:A capacitor is a device for storing energy. When we connect a battery across the two plates of a capacitor, the current charges the capacitor, leading to an accumulation of charges on opposite plates of the capacitor. As charges accumulate, the potential difference gradually increases across the two plates. Energy stored in a capacitor is the electrical potential energy which depends on the voltage and the capacitance of the capacitor.

Formula Used:
Energy stored in the capacitor is \[\dfrac{1}{2}C{V^2}\].
Here C is the capacitance and V is the voltage. By substituting both the value of capacitance and voltage (mentioned in the question), we can determine the value for energy stored in the capacitor. Both the values of capacitance (C) and voltage (V) are already given in the question.

Complete step by step answer:
As we know value of capacitance is given-\[16pF\]
And the voltage supply is \[70V\].
For finding the energy stored in the capacitor we can use the following formula-
\[E = \dfrac{1}{2}C{V^2}\]----- (1)
Here E is the energy stored in the capacitor (that we want to calculate) and C is the capacitance and V is the voltage supply.

Now start substituting all the given values in equation (1),
We get-
\[E = \dfrac{1}{2} \times 16 \times {10^{ - 12}} \times {\left( {80} \right)^2} \\
\therefore E = 5.6 \times {10^{ - 8}}J\]
So, this much energy is stored in the capacitor.

Hence option B is correct.

Note:Basically, capacitor is a device which stores electrical energy in the form of electrical charge that accumulated on their plates.Energy stored in the capacitor is electrostatic potential energy and it is related to the charge Q and voltage V between the capacitor plates. If a capacitor is charged then there will be an electrical field between its plates.