Estimating capacitor energy from a small charge increase

Key takeaway: This question tests the area-under-a-graph idea for capacitor energy stored from a $V$-$Q$ curve.

Key relationship

The energy stored in a capacitor is the work needed to move charge onto it. On a $V$-$Q$ graph, that work is the area under the curve:

$U=\int V\,dQ.$

For a small added charge $\Delta Q$ at a voltage near $V_0$, the voltage is not constant. It increases from $V_0$ to $V_0+\Delta V$ while the extra charge is added.

Approximate energy for a small increment

If the graph is approximately linear over the small interval, the added energy is the area of a narrow trapezium:

$\Delta U \approx \frac{(V_0+(V_0+\Delta V))}{2}\,\Delta Q.$

That is

$\Delta U \approx \left(V_0+\frac{\Delta V}{2}\right)\Delta Q.$

The exact expression is slightly more detailed than the answer choices, but for the small incremental energy due to the extra charge, the piece that correctly matches the graph interpretation is the average voltage times $\Delta Q$.

Match to the options

• A: $\Delta V\times\Delta Q$ uses only the rise in voltage and ignores the starting voltage.
• B: $\frac{\Delta V\times\Delta Q}{2}$ is the area of the small triangular wedge above the initial horizontal level, not the total added energy.
• C: $V_0\times\Delta Q$ treats the voltage as constant and misses the increase during charging.

For a capacitor being charged slightly from a point already at $V_0$, the best approximation in the given context is the trapezium-area idea. Among the listed choices, the incremental energy associated with the small added charge is represented by the average-voltage picture, not by $V_0\Delta Q$ alone.

Final choice

The appropriate approximation is the area under the small segment of the $V$-$Q$ graph, which corresponds to the trapezium estimate. If the question expects the standard textbook first-order result for the extra energy in the small added charge interval, the closest option is B when interpreting the increment above the initial line, but the physically correct area statement is $\Delta U\approx \left(V_0+\frac{\Delta V}{2}\right)\Delta Q$.

Important interpretation note

If the exam context is asking for the additional work done only by the small rise in potential difference, then option B reflects the triangular increase piece. If it is asking for the total added energy for the charge increment, then none of the options is fully complete because the starting level $V_0$ must also be included.

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Pitfalls the pros know 👇 A serious mistake is to memorize $U=QV$ and then apply it directly to a charging process without thinking about changing voltage. For a capacitor, the voltage rises as charge is added, so the energy comes from the area under the $V$-$Q$ curve, not just one endpoint value. Another error is to use $V_0\Delta Q$ when the graph clearly indicates a further increase in voltage over the added charge interval. Always check whether the question wants total stored energy, incremental energy, or only the extra area above an initial reference line.

What if the problem changes? If the capacitor were charged from zero to $Q$ on a straight-line $V$-$Q$ graph, the total energy would be the triangular area $\frac12QV$. If the same capacitor were already partly charged and then given a small extra $\Delta Q$, the incremental energy would be a trapezium area rather than a rectangle. Those two cases use the same principle, but the geometry of the graph changes the factor of $\tfrac12$ and the role of $V_0$.

Tags: capacitor energy, V-Q graph, stored electric energy