Electric potential and capacitance

What you will understand

For NEET UG students: decide the signs of potential-energy change and field work, then choose the quantity that stays fixed when a capacitor changes.

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Quick answer: write the constraint before a formula

For a test charge moved from A to B, $\Delta V=V_B-V_A$, $\Delta U=q\Delta V$ and $W_\mathrm{field}=-\Delta U$. For a linear capacitor, use $Q=CV$ and $U=Q^2/(2C)=CV^2/2$. Here $Q$ is the magnitude of the free charge on either plate and $V$ is the voltage magnitude between the plates. A connected ideal battery fixes $V$; an isolated disconnected capacitor with no leakage fixes $Q$. Those are different experiments.

Read the potential drop before the work sign

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Know what changes capacitance

For ideal parallel plates with area $A$, gap $d$ and a linear homogeneous dielectric completely filling that gap, $C=K\epsilon_0A/d$. Neglect fringing. $K$ is relative permittivity. Capacitance depends on geometry and material in this model, not on whichever $Q$ and $V$ happen to be present. Inserting $K=2$ at unchanged geometry doubles $C$; it does not by itself tell you whether $Q$ or $V$ doubles.

Compare the disconnected and connected experiments

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Common mistakes: identify the omitted condition

Wrong: “Dielectric always decreases capacitor energy.” Correct: it decreases energy for the isolated fixed-charge experiment but increases it at fixed voltage. Wrong: “Connected means charge stays fixed.” Correct: charge can flow through the battery leads. Wrong: use $E=\Delta V/d$ without signs for an arbitrary direction. Correct: the field component is $-dV/dx$; a magnitude formula needs the appropriate direction and uniform-field assumption. Wrong: external work always equals $q\Delta V$. Correct: if kinetic energy changes and only electric and external work act, $W_\mathrm{ext}=\Delta K+\Delta U$.

Try it: one charge, two signs

Move $q=+3\,\mu\mathrm C$ from A at $20\,\mathrm V$ to B at $-10\,\mathrm V$. Find $\Delta U$ and work by the field. What would quasi-static external work be if there is no other work?

Try it: double the gap after disconnecting

A vacuum parallel-plate capacitor is initially $3\,\mu\mathrm F$ at $4\,\mathrm V$. Disconnect and isolate it with no leakage, then double the plate gap while retaining the plate area. Neglect fringing. Find its new capacitance, charge, voltage and stored energy.

NEET recall: make a constraint table

Rebuild three rows from memory: connected battery → fixed voltage; disconnected and isolated with no leakage → fixed charge; full homogeneous dielectric at unchanged geometry → capacitance multiplied by $K$. Then derive the remaining quantities with $Q=CV$ and one energy expression. Use the diagram only after you have predicted whether voltage, charge and energy rise or fall.

Come back tomorrow: solve the other experiment

Re-solve the gap-doubling self-check while keeping the ideal $4\,\mathrm V$ battery connected. Predict the changes first. The result is $C=1.5\,\mu\mathrm F$, $Q=6\,\mu\mathrm C$ and $U=12\,\mu\mathrm J$. Explain why these differ from the disconnected answers. Continue with broad chapter practice below; these links are not a dedicated dielectric set or a promise of every JEE question format.

Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Numerical examples, self-checks and diagrams are original. Human subject review is not claimed.

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