Which expression correctly gives the energy stored in a capacitor in terms of its capacitance C and voltage V?

Study for the Electrostatics Test. Enhance your understanding with interactive questions, detailed explanations, and comprehensive review. Prepare for success!

Multiple Choice

Which expression correctly gives the energy stored in a capacitor in terms of its capacitance C and voltage V?

Explanation:
When a capacitor is charged, the work done to move charge into it builds up the stored energy. The differential work is dW = V dq, because each small amount of charge dq is moved against the instantaneous voltage V. For a capacitor, the voltage is related to the charge by V = Q/C, so dW = (q/C) dq. Integrating from zero to the final charge Q gives U = ∫0^Q (q/C) dq = Q^2/(2C). Since Q = CV, this becomes U = (1/2) C V^2. This form is equivalent to U = (1/2) QV, since Q = CV. The other expressions aren’t correct in general: U = QV would only be valid if the voltage stayed constant while charging, which isn’t the case for a capacitor, and U = (1/2) Q^2 V has the wrong dependence on Q and V (it doesn’t simplify to the correct energy with the right units).

When a capacitor is charged, the work done to move charge into it builds up the stored energy. The differential work is dW = V dq, because each small amount of charge dq is moved against the instantaneous voltage V. For a capacitor, the voltage is related to the charge by V = Q/C, so dW = (q/C) dq. Integrating from zero to the final charge Q gives U = ∫0^Q (q/C) dq = Q^2/(2C). Since Q = CV, this becomes U = (1/2) C V^2.

This form is equivalent to U = (1/2) QV, since Q = CV. The other expressions aren’t correct in general: U = QV would only be valid if the voltage stayed constant while charging, which isn’t the case for a capacitor, and U = (1/2) Q^2 V has the wrong dependence on Q and V (it doesn’t simplify to the correct energy with the right units).

Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy