If potential difference across two plates of a parallel plate capacitor is doubled then find the energy stored in it ?
The correct answer and explanation is:
Let’s analyze the problem step-by-step:
Problem:
If the potential difference (voltage) across two plates of a parallel plate capacitor is doubled, what happens to the energy stored in the capacitor?
Given:
- Initial potential difference = VV
- New potential difference = 2V2V
- Capacitance = CC (constant, since the physical setup of the capacitor — plate area and distance between plates — doesn’t change)
Relevant Formula:
The energy stored in a capacitor is given by: U=12CV2U = \frac{1}{2} C V^2
where
- UU = energy stored
- CC = capacitance
- VV = potential difference (voltage) across the plates
Solution:
- Initial energy stored is:
U1=12CV2U_1 = \frac{1}{2} C V^2
- When the voltage is doubled, V→2VV \to 2V, the new energy stored is:
U2=12C(2V)2=12C×4V2=4×12CV2=4U1U_2 = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \times 4V^2 = 4 \times \frac{1}{2} C V^2 = 4 U_1
Answer:
When the potential difference across the plates is doubled, the energy stored in the capacitor increases by a factor of 4.
Explanation (300 words):
A parallel plate capacitor stores electrical energy in the electric field between its plates. The amount of energy stored depends on two key factors: the capacitance CC, which is determined by the physical characteristics of the capacitor (plate area, distance between plates, and the dielectric material), and the voltage VV applied across its plates.
The energy stored in a capacitor is given by the formula U=12CV2U = \frac{1}{2} C V^2. This formula indicates that the energy stored is proportional to the square of the voltage. This quadratic relationship is important because even small changes in voltage cause larger changes in stored energy.
When the voltage is doubled from VV to 2V2V, the energy stored becomes: U2=12C(2V)2=12C×4V2=4×U1U_2 = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \times 4 V^2 = 4 \times U_1
This means that doubling the voltage increases the energy stored by four times.
Physically, this happens because increasing voltage increases the electric field strength between the plates, which stores more energy per unit volume in the capacitor. Since the capacitance remains constant (no change in plate area or distance), the energy change comes entirely from the voltage increase.
This principle is important in applications where capacitors are used to store energy — for example, in flash photography or pulsed power applications. It shows that careful control of voltage is critical since energy scales with the square of voltage, impacting how much power can be delivered and how safe the capacitor operates.