A parallel-plate capacitor connected to an ideal battery has charge \(+Q\) on its top plate. The energy stored in the capacitor is \(U_{c}\). While the capacitor remains connected to the battery, the separation between the two plates is doubled. Which of the following gives the new charge on the top plate and the new energy stored in the capacitor?
Charge Potential Energy
(A) \(+2Q\) \(2U_{c}\)
(B) \(+2Q\) \(\dfrac{U_{c}}{2}\)
(C) \(+Q\) \(U_{c}\)
(D) \(+\dfrac{Q}{2}\) \(2U_{c}\)
(E) \(+\dfrac{Q}{2}\) \(\dfrac{U_{c}}{2}\)
Correct Answer: \( \boxed{\mathrm{E}} \)
Since the capacitor remains connected to an ideal battery, the potential difference remains constant.
For a parallel-plate capacitor,
\( C=\dfrac{\varepsilon_{0}A}{d} \)
Doubling the plate separation doubles \(d\), so the capacitance becomes
\( C’=\dfrac{C}{2} \)
The charge is
\( Q=CV \)
Since \(V\) is constant,
\( Q’=C’V=\dfrac{Q}{2} \)
The energy stored is
\( U=\dfrac{1}{2}CV^{2} \)
Again, because \(V\) remains constant,
\( U’=\dfrac{1}{2}C’V^{2}=\dfrac{U_{c}}{2} \)
Therefore, the new charge is \(+\dfrac{Q}{2}\) and the new energy stored is \(\dfrac{U_{c}}{2}\).
Therefore, the correct answer is (E).
A parallel-plate capacitor is filled with air. Each plate has area \(10~\mathrm{cm^2}\). The separation between the plates is \(2.0~\mathrm{cm}\). The top plate stores \(+200~\mathrm{nC}\) of charge, and the bottom plate stores \(-200~\mathrm{nC}\) of charge.
The original capacitor is again filled with air. Now the plates are replaced with new plates of area \(20~\mathrm{cm^2}\) but the same charge and separation. How is the potential difference across the plates affected?
Correct Answer: \( \boxed{\mathrm{B}} \)
The capacitance of a parallel-plate capacitor is
\( C=\dfrac{\varepsilon_0A}{d} \)
Since the plate area doubles while the plate separation remains the same, the capacitance doubles:
\( C’=2C \)
The charge on the capacitor remains unchanged:
\( Q=\mathrm{constant} \)
Using
\( V=\dfrac{Q}{C} \)
the new potential difference is
\( V’=\dfrac{Q}{2C}=\dfrac{V}{2} \)
Therefore, doubling the plate area doubles the capacitance and reduces the potential difference to one-half of its original value.
Hence, the correct answer is (B).
Four identical capacitors of capacitance \(C\) are connected as illustrated above. What is their equivalent capacitance?
Correct Answer: \( \boxed{\mathrm{C}} \)
The two capacitors on the right are connected in parallel, so their equivalent capacitance is
\( C_{\mathrm{p}}=C+C=2C \)
This equivalent capacitor is in series with the capacitor in the top branch:
\( C_{\mathrm{s}}=\dfrac{C(2C)}{C+2C}=\dfrac{2C}{3} \)
Finally, this series combination is in parallel with the left capacitor, giving
\( C_{\mathrm{eq}}=C+\dfrac{2C}{3}=\dfrac{5C}{3} \)
Therefore, the equivalent capacitance is
\( \boxed{C_{\mathrm{eq}}=\dfrac{5C}{3}} \)
Hence, the correct answer is (C).
