A parallel-plate capacitor has a dielectric material between the plates with dielectric constant \(K\). The capacitor is connected to a variable resistor \(R\) and a power supply of potential difference \(V\). Each plate has cross-sectional area \(A\), and the plates are separated by a distance \(d\).
Which of the following changes could increase the capacitance and decrease the amount of charge stored on the capacitor?
Correct Answer: \( \boxed{\mathrm{B}} \)
The capacitance of a parallel-plate capacitor is
\(C=\dfrac{K\varepsilon_0A}{d}\).
Therefore, capacitance increases if \(K\) or \(A\) increases, or if the plate separation \(d\) decreases.
The charge stored on the capacitor is
\(Q=CV\).
The resistor \(R\) affects only the charging rate (time constant) and does not affect the final capacitance or stored charge.
In choice (B), decreasing \(d\) increases the capacitance, while decreasing the battery voltage \(V\) reduces the stored charge. Since \(Q=CV\), a sufficiently smaller \(V\) results in a smaller \(Q\) even though \(C\) has increased.
Therefore, the correct answer is (B).
The four capacitors in the combination illustrated above each have capacitance \(C\). If all the capacitors are then filled with a dielectric having dielectric constant \(K=2\), what is the new total capacitance of the combination?
Correct Answer: \( \boxed{\mathrm{B}} \)
First, determine the initial equivalent capacitance.
The two capacitors inside the square are connected in parallel, so
\( C_{\mathrm{p}}=C+C=2C \)
The circuit is then reduced to three capacitors in series:
\( C,\;2C,\;C \)
For capacitors in series,
\( \dfrac{1}{C_{\mathrm{eq}}}=\dfrac{1}{C}+\dfrac{1}{2C}+\dfrac{1}{C}=\dfrac{5}{2C} \)
Therefore, the initial equivalent capacitance is
\( C_{\mathrm{eq,initial}}=\dfrac{2C}{5} \)
When every capacitor is completely filled with a dielectric of dielectric constant \(K=2\), each individual capacitance doubles:
\( C’=KC=2C \)
Since every capacitor is multiplied by the same factor, the equivalent capacitance is also multiplied by the same factor:
\( C_{\mathrm{eq,new}}=K\,C_{\mathrm{eq,initial}} \)
Substituting \(K=2\),
\( C_{\mathrm{eq,new}}=2\left(\dfrac{2C}{5}\right)=\dfrac{4C}{5} \)
Therefore, the new equivalent capacitance of the combination is
\( \boxed{C_{\mathrm{eq,new}}=\dfrac{4C}{5}} \)
Hence, the correct answer is (B).
Four parallel-plate capacitors all have the same plate area and have the plate separations shown above. Both capacitors \(A\) and \(B\) have air between the plates, while the space between the plates of both capacitors \(C\) and \(D\) is filled with a dielectric slab of dielectric constant \(K=2\). Which of the following correctly ranks the capacitors in order of their capacitance from largest to smallest?
Correct Answer: \( \boxed{\mathrm{C}} \)
The capacitance of a parallel-plate capacitor is
\( C=\dfrac{K\varepsilon_0A}{d} \)
Since all four capacitors have the same plate area \(A\), compare only the dielectric constant \(K\) and the plate separation \(d\).
Their capacitances are:
\( C_A=\dfrac{\varepsilon_0A}{d} \)
\( C_B=\dfrac{\varepsilon_0A}{2d}=\dfrac{1}{2}C_A \)
\( C_C=\dfrac{2\varepsilon_0A}{d}=2C_A \)
\( C_D=\dfrac{2\varepsilon_0A}{2d}=C_A \)
Thus, the ranking from largest to smallest is
\( C>C_A=D>B \)
or
\( \boxed{C>(A=D)>B} \)
Hence, the correct answer is (C).
