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 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
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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!
Updated On: Jun 3, 2026
- figure A
- figure B
- figure C
- figure D
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The Correct Option is A
Solution and Explanation
Step 1: Concept
When a charged capacitor is disconnected from its charging source, its total electric charge ($Q$) is trapped and stays constant. The potential difference across the plates is governed by $V = \frac{Q}{C}$.
Step 2: Meaning
Pulling out a dielectric slab reduces the effective capacitance $C$ of the system. As a result, the potential difference $V$ must increase as the slab is extracted.
Step 3: Analysis
Let the length of the plates be $L$ and the length of the dielectric pulled out be $x$. The capacitance can be modeled as two parallel capacitors in combination: one part filled with air of length $x$, and one part filled with the dielectric of remaining length $(L-x)$. This combination yields a capacitance that decreases as a function of $x$. For relatively small extractions or linear approximations of fractional change, the voltage $V(x) = \frac{Q}{C(x)}$ shows a steady, monotonic upward trend with respect to displacement $x$, starting from an initial non-zero baseline potential $V_0$.
Step 4: Conclusion
This corresponding relationship matches a graph showing a linear increase in voltage with respect to displacement $x$.
Final Answer: (A)
When a charged capacitor is disconnected from its charging source, its total electric charge ($Q$) is trapped and stays constant. The potential difference across the plates is governed by $V = \frac{Q}{C}$.
Step 2: Meaning
Pulling out a dielectric slab reduces the effective capacitance $C$ of the system. As a result, the potential difference $V$ must increase as the slab is extracted.
Step 3: Analysis
Let the length of the plates be $L$ and the length of the dielectric pulled out be $x$. The capacitance can be modeled as two parallel capacitors in combination: one part filled with air of length $x$, and one part filled with the dielectric of remaining length $(L-x)$. This combination yields a capacitance that decreases as a function of $x$. For relatively small extractions or linear approximations of fractional change, the voltage $V(x) = \frac{Q}{C(x)}$ shows a steady, monotonic upward trend with respect to displacement $x$, starting from an initial non-zero baseline potential $V_0$.
Step 4: Conclusion
This corresponding relationship matches a graph showing a linear increase in voltage with respect to displacement $x$.
Final Answer: (A)
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