Question 551: Electrostatics & Capacitance

Question

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles, then

Accepted Answer

For a parallel plate capacitor, $$\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$$
When the charging battery is disconnected and $$\mathrm{d}$$ is increased then
(a) Q remains constant
(b) $$\mathrm{C} \propto \frac{1}{\mathrm{~d}}$$ hence if $$\mathrm{d}$$ is increased, $$\mathrm{C}$$ is decreased.
(c) $$\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$$ or $$\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$$ if $$\mathrm{C}$$ is decreased, $$\mathrm{V}$$ will increase
(d) $$\mathrm{E}=\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}$$ if $$\mathrm{C}$$ is decreased, then $$\mathrm{E}$$ will increase
Thus (a), (b) and (d) are wrong. Only (c) is correct.

Answer

the capacitance of the capacitor increases

Answer

the charge on the capacitor decreases

Answer

the voltage across the capacitor increases

Answer

the energy stored in the capacitor decreases

Electrostatics & Capacitance

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles, then

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