Question

Prove that energy stored in the parallel plate capacitor is given by 1/2cv2

Answer 1

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Answer 2

Answer:

The energy stored in the parallel plate capacitor is $\dfrac{1}{2} CV^{2}$.

Explanation:

In a parallel plate capacitor, when the parallel plates are connected across a battery, the plates are charged and an electric field is established between them.    

• The maximum amount of charge that can be acquired by a capacitor is proportional to the voltage.

        $Q\propto V\implies Q=CV$    

• where C, the proportionality constant, called capacitance.
• The capacitance depends on the geometry of the capacitor, therefore

       $C=\dfrac{\varepsilon A}{d}$

       where A is the area of a plate, d is the distance between the plates, and  $\varepsilon$ is the electric permittivity.

• Energy stored in the capacitor is the work done to move a charge Q through a potential difference V.  
• The electric potential energy stored is given by the charge times the average potential difference.
• $U=\dfrac{1}{2} QV=\dfrac{1}{2} (CV)V=\dfrac{1}{2} CV^{2}$

Therefore, the energy stored in the parallel plate capacitor is $\dfrac{1}{2} CV^{2}$.