Question

A parallel plate capacitor with dielectric is charged completely and the battery is then disconnected. Now if the dielectric is slowly pulled out of the capacitor then the variation of the potential (V) of the capacitor with respect to the length (x) of the dielectric pulled out is represented by

• A. figure A
• B. figure B
• C. figure C
• D. figure D

Hint:

Battery disconnected $\implies Q$ is constant. Pulling out dielectric $\implies C$ decreases. Since $V = Q/C$, decreasing $C$ forces the potential $V$ to go up!

Answer 1

The Correct Option is A

Step 1: Charge stays fixed.
After the battery is removed, no charge can flow in or out. So the charge $Q$ on the plates is constant.

Step 2: Voltage depends on capacitance.
The plate voltage is $V = \dfrac{Q}{C}$. If $C$ changes, $V$ must change since $Q$ is fixed.

Step 3: Pulling out the slab lowers $C$.
A dielectric raises capacitance, so removing it lowers $C$. As more slab comes out, $C$ keeps falling.

Step 4: So voltage rises.
Since $V = \dfrac{Q}{C}$ and $C$ goes down, $V$ goes up as the slab is pulled out.

Step 5: Find the shape of the rise.
Modelling the slab-out part and slab-in part as two parallel capacitors, $C(x)$ falls steadily with the pulled-out length $x$, so $V(x)$ climbs steadily from a non-zero start value.

Step 6: Match the graph.
The right graph shows voltage starting at a positive value and increasing with $x$. This matches option 1.
\[ \boxed{V \text{ increases as the dielectric is pulled out}} \]