Work done in charging a capacitor is ____________

Question

Examoogle · Accepted Answer

$0.5QV$

Explanation: The work done in charging a capacitor is equal to the electrostatic potential energy stored in its electric field. Since the potential across a capacitor increases linearly as charge is added, the work done is represented by the average of the initial and final energy states.

Working Principle: When charging a capacitor from $0$ to a charge $Q$ at a potential $V$, the instantaneous potential is $v = q/C$. The total work $W$ is the integral of $v dq$ from $0$ to $Q$, which yields $W = \int_0^Q \frac{q}{C} dq = \frac{1}{2} \frac{Q^2}{C}$. Substituting $Q = CV$, we obtain $W = 0.5QV$.

Key Formulas: $W = \frac{1}{2}QV$; $W = \frac{1}{2}CV^2$; $W = \frac{Q^2}{2C}$

Key Points: Energy is stored in the electric field between the capacitor plates.; The factor of 0.5 accounts for the fact that the charging process starts at zero voltage.; The unit of work done is Joules (J).

Advantages: Efficient energy storage for power electronics.; Rapid charge and discharge capability.

Disadvantages: Limited energy density compared to batteries.; Self-discharge over long periods.

Applications: Filtering in power supply circuits.; Energy storage in camera flashes and defibrillators.

Additional Info: Option A (QV) represents the work done by a source in moving charge Q through a potential V in a resistive circuit, not a capacitor.. The area under the V-Q graph (a triangle) gives the value of $0.5QV$.

Answer: B is correct — The work done to charge a capacitor is $0.5QV$ because the potential increases linearly from $0$ to $V$ during the charging process.

Exam Tip: Remember that in a resistive circuit, power is $VI$, but for capacitive energy storage, always include the $0.5$ factor due to the charging profile.

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