301. A parallel-plate capacitor is fully filled with a dielectric of constant \(K\). Which row correctly compares the electric field after insertion for isolated and battery-connected cases?
| Row | Isolated capacitor | Battery-connected capacitor |
|---|
| P | Field becomes \(\frac{E_0}{K}\) | Field remains \(\frac{V_0}{d}\) |
| Q | Field remains \(E_0\) | Field becomes \(\frac{E_0}{K}\) |
| R | Field becomes \(KE_0\) | Field becomes \(KE_0\) |
| S | Field becomes zero | Field becomes zero |
ⓐ. Row Q
ⓑ. Row S
ⓒ. Row R
ⓓ. Row P
Correct Answer: Row P
Explanation: In an isolated capacitor, the free charge \(Q\) remains fixed after the dielectric is inserted. The dielectric polarization reduces the electric field for the same free charge, so \(E\) becomes \(\frac{E_0}{K}\). In a battery-connected capacitor, the battery keeps the potential difference \(V_0\) fixed. Since the plate separation \(d\) is unchanged, the field magnitude remains \(E=\frac{V_0}{d}\). Extra charge flows from the battery to maintain this field while capacitance increases. The important contrast is fixed \(Q\) versus fixed \(V\). A dielectric reduces field only directly in the fixed-free-charge situation.
302. A parallel-plate capacitor is fully filled with a dielectric of permittivity \(\varepsilon=K\varepsilon_0\). If the electric field inside the dielectric is \(E\), what is the energy density?
ⓐ. \(u=\frac{1}{2}\varepsilon E^2\)
ⓑ. \(u=\varepsilon E\)
ⓒ. \(u=\frac{1}{2}\varepsilon_0\frac{E^2}{K}\)
ⓓ. \(u=\frac{E^2}{2\varepsilon}\)
Correct Answer: \(u=\frac{1}{2}\varepsilon E^2\)
Explanation: The energy density of an electric field in a dielectric medium is \(u=\frac{1}{2}\varepsilon E^2\). Here \(\varepsilon=K\varepsilon_0\), so the medium response is included through the permittivity. The formula gives energy per unit volume, measured in \(\text{J m}^{-3}\). The field appears as \(E^2\), so energy density is independent of the sign or direction of the field. The expression \(\frac{E^2}{2\varepsilon}\) has the wrong dependence on permittivity for this capacitor-energy context. The safe distinction is: use \(\varepsilon_0\) in vacuum and \(\varepsilon=K\varepsilon_0\) in a dielectric.
303. A capacitor is connected to a battery and has plate separation \(d\). A dielectric is inserted fully while the battery remains connected. Which statement best describes charge, field, and energy in the capacitor?
ⓐ. \(Q\) remains fixed, \(E\) decreases to \(\frac{E_0}{K}\), and stored energy decreases
ⓑ. \(Q\) increases, \(E\) remains fixed by \(\frac{V}{d}\), and stored energy increases
ⓒ. \(Q\) increases, \(E\) decreases to \(\frac{E_0}{K}\), and stored energy remains unchanged
ⓓ. \(Q\) decreases, \(E\) becomes zero, and stored energy becomes zero
Correct Answer: \(Q\) increases, \(E\) remains fixed by \(\frac{V}{d}\), and stored energy increases
Explanation: A connected battery fixes the potential difference \(V\) across the plates. Since \(d\) is unchanged, the field between ideal parallel plates remains \(E=\frac{V}{d}\). The dielectric increases capacitance from \(C_0\) to \(KC_0\). Because \(Q=CV\), extra charge flows from the battery and \(Q\) increases. The stored energy is \(U=\frac{1}{2}CV^2\), so it increases when \(C\) increases at fixed \(V\). The common mistake is to apply the isolated-capacitor field reduction rule even though the battery is still connected.
304. Two capacitors \(C_1=2.0\,\mu\text{F}\) and \(C_2=8.0\,\mu\text{F}\) are connected in series across \(100\,\text{V}\). Which capacitor stores more energy, and what is the ratio \(U_1:U_2\)?
ⓐ. The \(2.0\,\mu\text{F}\) capacitor stores more energy, with \(U_1:U_2=4:1\)
ⓑ. The \(8.0\,\mu\text{F}\) capacitor stores more energy, with \(U_1:U_2=1:4\)
ⓒ. The \(2.0\,\mu\text{F}\) capacitor stores more energy, with \(U_1:U_2=2:1\)
ⓓ. Both store equal energy, with \(U_1:U_2=1:1\)
Correct Answer: The \(2.0\,\mu\text{F}\) capacitor stores more energy, with \(U_1:U_2=4:1\)
Explanation: \( \textbf{Series condition:} \) Both capacitors carry the same charge \(Q\).
\( \textbf{Energy at fixed charge:} \)
\[
U=\frac{Q^2}{2C}
\]
\( \textbf{Energy ratio:} \)
\[
U_1:U_2=\frac{1}{C_1}:\frac{1}{C_2}
\]
\( \textbf{Substitution:} \)
\[
U_1:U_2=\frac{1}{2.0}:\frac{1}{8.0}
\]
\( \textbf{Simplification:} \)
\[
U_1:U_2=4:1
\]
\( \textbf{Physical meaning:} \) The smaller capacitor has larger voltage and stores more energy in a series combination.
\( \textbf{Final answer:} \) The \(2.0\,\mu\text{F}\) capacitor stores more energy, and \(U_1:U_2=4:1\).
305. A \(3.0\,\mu\text{F}\) capacitor charged to \(90\,\text{V}\) is connected in parallel with an uncharged \(6.0\,\mu\text{F}\) capacitor. What is the energy lost during redistribution?
ⓐ. \(8.1\times10^{-3}\,\text{J}\)
ⓑ. \(0\,\text{J}\)
ⓒ. \(2.43\times10^{-2}\,\text{J}\)
ⓓ. \(1.62\times10^{-2}\,\text{J}\)
Correct Answer: \(8.1\times10^{-3}\,\text{J}\)
Explanation: \( \textbf{Initial charge:} \)
\[
Q_i=C_1V_1=(3.0\,\mu\text{F})(90\,\text{V})=270\,\mu\text{C}
\]
\( \textbf{Final common potential:} \)
\[
V_f=\frac{Q_i}{C_1+C_2}
\]
\( \textbf{Substitution:} \)
\[
V_f=\frac{270\,\mu\text{C}}{3.0\,\mu\text{F}+6.0\,\mu\text{F}}=30\,\text{V}
\]
\( \textbf{Initial energy:} \)
\[
U_i=\frac{1}{2}C_1V_1^2
\]
\( \textbf{Calculation of \(U_i\):} \)
\[
U_i=\frac{1}{2}(3.0\times10^{-6})(90)^2=1.215\times10^{-2}\,\text{J}
\]
\( \textbf{Final energy:} \)
\[
U_f=\frac{1}{2}(C_1+C_2)V_f^2
\]
\( \textbf{Calculation of \(U_f\):} \)
\[
U_f=\frac{1}{2}(9.0\times10^{-6})(30)^2=4.05\times10^{-3}\,\text{J}
\]
\( \textbf{Energy lost:} \)
\[
U_i-U_f=1.215\times10^{-2}-4.05\times10^{-3}=8.10\times10^{-3}\,\text{J}
\]
\( \textbf{Final answer:} \) The energy lost is \(8.1\times10^{-3}\,\text{J}\).
306. Two capacitors \(C_1\) and \(C_2\) are connected after being charged to different potentials. Which condition makes the final common potential zero?
ⓐ. The two capacitances are connected in series after charging
ⓑ. \(C_1=C_2\) only, regardless of polarity and voltage
ⓒ. \(V_1=V_2\) only, regardless of capacitance and polarity
ⓓ. Equal and opposite initial charges on the joined plates
Correct Answer: Equal and opposite initial charges on the joined plates
Explanation: The final common potential after parallel connection depends on conserved algebraic charge. The initial algebraic charge contribution from a capacitor is \(CV\), with sign decided by the joined polarity. If opposite polarities are connected and \(C_1V_1=C_2V_2\), the algebraic total charge is zero. The final capacitance is \(C_1+C_2\), so \(V_f=\frac{0}{C_1+C_2}=0\). Equal capacitance alone is not sufficient unless the voltage magnitudes and polarities also produce cancellation. The important sign-related point is to conserve algebraic charge, not just add charge magnitudes.
307. A Van de Graaff dome is approximated as an isolated conducting sphere of radius \(0.50\,\text{m}\). If air breaks down when the surface electric field reaches \(3.0\times10^6\,\text{V m}^{-1}\), estimate the maximum potential of the dome.
ⓐ. \(3.0\times10^6\,\text{V}\)
ⓑ. \(6.0\times10^6\,\text{V}\)
ⓒ. \(1.5\times10^6\,\text{V}\)
ⓓ. \(7.5\times10^5\,\text{V}\)
Correct Answer: \(1.5\times10^6\,\text{V}\)
Explanation: \( \textbf{Given:} \) Sphere radius \(R=0.50\,\text{m}\), and maximum surface field \(E_{\max}=3.0\times10^6\,\text{V m}^{-1}\).
\( \textbf{Sphere relations:} \) For a conducting sphere,
\[
V=k\frac{Q}{R}
\]
and
\[
E=k\frac{Q}{R^2}
\]
\( \textbf{Connect \(V\) and \(E\):} \)
\[
V=ER
\]
\( \textbf{Substitution:} \)
\[
V_{\max}=(3.0\times10^6)(0.50)
\]
\( \textbf{Calculation:} \)
\[
V_{\max}=1.5\times10^6\,\text{V}
\]
\( \textbf{Physical meaning:} \) A larger dome can reach higher potential for the same breakdown field.
\( \textbf{Final answer:} \) The maximum potential is about \(1.5\times10^6\,\text{V}\).
308. A conducting sphere used as a Van de Graaff dome has capacitance \(C=4\pi\varepsilon_0R\). If its radius is doubled while the stored charge remains the same, what happens to its potential?
ⓐ. It becomes half
ⓑ. It becomes four times
ⓒ. It doubles
ⓓ. It remains unchanged
Correct Answer: It becomes half
Explanation: The potential of an isolated conducting sphere can be written as \(V=\frac{Q}{C}\). Its capacitance is \(C=4\pi\varepsilon_0R\), so \(C\propto R\). If the radius is doubled, the capacitance doubles. With stored charge \(Q\) unchanged, \(V=\frac{Q}{C}\) becomes half. This is why a larger Van de Graaff dome can hold more charge for the same potential or lower surface field for comparable charge. The conductor remains equipotential throughout. A common proportionality mistake is to think larger size always means larger potential for the same charge; actually, larger capacitance lowers \(V\) at fixed \(Q\).
309. A parallel-plate capacitor in vacuum has plate area \(A\), separation \(d\), and electric field \(E\). Which expression gives the total energy stored using energy density?
ⓐ. \(U=\varepsilon_0EAd\)
ⓑ. \(U=\frac{1}{2}\varepsilon_0E A\)
ⓒ. \(U=\frac{1}{2}\varepsilon_0E^2Ad\)
ⓓ. \(U=\frac{1}{2}\frac{E^2}{\varepsilon_0}Ad\)
Correct Answer: \(U=\frac{1}{2}\varepsilon_0E^2Ad\)
Explanation: \( \textbf{Energy density in vacuum:} \)
\[
u=\frac{1}{2}\varepsilon_0E^2
\]
\( \textbf{Volume between plates:} \)
\[
\text{Volume}=Ad
\]
\( \textbf{Total energy from density:} \)
\[
U=u(\text{Volume})
\]
\( \textbf{Substitution:} \)
\[
U=\left(\frac{1}{2}\varepsilon_0E^2\right)(Ad)
\]
\( \textbf{Final expression:} \)
\[
U=\frac{1}{2}\varepsilon_0E^2Ad
\]
\( \textbf{Consistency check:} \) Using \(C=\frac{\varepsilon_0A}{d}\) and \(V=Ed\), \(\frac{1}{2}CV^2\) gives the same result.
\( \textbf{Final answer:} \) \(U=\frac{1}{2}\varepsilon_0E^2Ad\).
310. A capacitor is charged by a battery and then disconnected. Its plate area is doubled by replacing the plates with larger identical-shape plates while charge and separation are kept unchanged. What happens to capacitance, voltage, and stored energy?
ⓐ. \(C\) remains unchanged, \(V\) becomes half, and \(U\) remains unchanged
ⓑ. \(C\) becomes half, \(V\) doubles, and \(U\) doubles
ⓒ. \(C\) doubles, \(V\) becomes half, and \(U\) becomes half
ⓓ. \(C\) doubles, \(V\) doubles, and \(U\) doubles
Correct Answer: \(C\) doubles, \(V\) becomes half, and \(U\) becomes half
Explanation: \( \textbf{Capacitance relation:} \)
\[
C=\frac{\varepsilon_0A}{d}
\]
\( \textbf{Area change:} \) If \(A\) is doubled while \(d\) is unchanged, capacitance doubles.
\( \textbf{Disconnected condition:} \) The charge \(Q\) remains fixed.
\( \textbf{Voltage relation:} \)
\[
V=\frac{Q}{C}
\]
\( \textbf{Voltage change:} \) If \(C\) doubles at fixed \(Q\), \(V\) becomes half.
\( \textbf{Energy relation for fixed charge:} \)
\[
U=\frac{Q^2}{2C}
\]
\( \textbf{Energy change:} \) If \(C\) doubles, \(U\) becomes half.
\( \textbf{Final answer:} \) \(C\) doubles, \(V\) becomes half, and \(U\) becomes half.
311. Two fixed charges \(+Q\) and \(+4Q\) are separated by distance \(a\). At the point between them where the electric field is zero, what is the electric potential?
ⓐ. \(0\)
ⓑ. \(\frac{9kQ}{a}\)
ⓒ. \(\frac{3kQ}{a}\)
ⓓ. \(\frac{6kQ}{a}\)
Correct Answer: \(\frac{9kQ}{a}\)
Explanation: \( \textbf{Given:} \) Charges \(+Q\) and \(+4Q\) are separated by \(a\).
\( \textbf{Field-zero condition:} \) For like charges, the zero-field point lies between them.
\( \textbf{Let distance from \(+Q\):} \) \(x\), so distance from \(+4Q\) is \(a-x\).
\( \textbf{Set field magnitudes equal:} \)
\[
k\frac{Q}{x^2}=k\frac{4Q}{(a-x)^2}
\]
\( \textbf{Simplify:} \)
\[
\frac{1}{x^2}=\frac{4}{(a-x)^2}
\]
\[
a-x=2x
\]
\[
x=\frac{a}{3}
\]
\( \textbf{Distance from \(+4Q\):} \)
\[
a-x=\frac{2a}{3}
\]
\( \textbf{Potential at that point:} \)
\[
V=k\frac{Q}{a/3}+k\frac{4Q}{2a/3}
\]
\( \textbf{Simplification:} \)
\[
V=\frac{3kQ}{a}+\frac{6kQ}{a}=\frac{9kQ}{a}
\]
\( \textbf{Final answer:} \) \(V=\frac{9kQ}{a}\), showing that zero electric field does not imply zero potential.
312. A short electric dipole has moment \(p\). A point \(P\) is at distance \(r\) from its centre and makes angle \(\theta\) with \(\vec{p}\). A charge \(-q\) is slowly brought from infinity to \(P\). What is the external work required?
ⓐ. \(-k\frac{qp\cos\theta}{r^2}\)
ⓑ. \(+k\frac{qp\cos\theta}{r^2}\)
ⓒ. \(-k\frac{qp\sin\theta}{r^2}\)
ⓓ. \(+k\frac{qp\cos\theta}{r^3}\)
Correct Answer: \(-k\frac{qp\cos\theta}{r^2}\)
Explanation: \( \textbf{Dipole potential:} \) At a far point,
\[
V=k\frac{p\cos\theta}{r^2}
\]
\( \textbf{Charge brought:} \) The charge is \(-q\).
\( \textbf{Potential energy in external potential:} \)
\[
U=q_{\text{test}}V
\]
\( \textbf{Substitution:} \)
\[
U=(-q)\left(k\frac{p\cos\theta}{r^2}\right)
\]
\( \textbf{Simplification:} \)
\[
U=-k\frac{qp\cos\theta}{r^2}
\]
\( \textbf{Slow movement condition:} \) External work equals the change in electrostatic potential energy.
\( \textbf{Initial reference:} \) At infinity, \(U_i=0\).
\( \textbf{Final external work:} \)
\[
W_{\text{ext}}=U_f-U_i=-k\frac{qp\cos\theta}{r^2}
\]
\( \textbf{Sign interpretation:} \) The sign depends on \(\cos\theta\); on the positive axial side, the negative charge is attracted and external work is negative.
\( \textbf{Final answer:} \) \(W_{\text{ext}}=-k\frac{qp\cos\theta}{r^2}\).
313. Four charges are placed at the corners of a square of side \(a\): \(+q\) at the top-left and bottom-right corners, and \(-q\) at the top-right and bottom-left corners. What is the total electrostatic potential energy of the system?
ⓐ. \(\frac{kq^2}{a}(-4+\sqrt{2})\)
ⓑ. \(\frac{kq^2}{a}(4+\sqrt{2})\)
ⓒ. \(\frac{kq^2}{a}(-4-\sqrt{2})\)
ⓓ. \(\frac{kq^2}{a}(4-\sqrt{2})\)
Correct Answer: \(\frac{kq^2}{a}(-4+\sqrt{2})\)
Explanation: \( \textbf{Pair counting:} \) Four charges form \(6\) distinct pairs.
\( \textbf{Side pairs:} \) Each side of the square joins unlike charges, so each side-pair contribution is
\[
-\frac{kq^2}{a}
\]
\( \textbf{Total side contribution:} \)
\[
U_{\text{sides}}=4\left(-\frac{kq^2}{a}\right)=-\frac{4kq^2}{a}
\]
\( \textbf{Diagonal pairs:} \) The two diagonals join like charges: one \(+q,+q\) pair and one \(-q,-q\) pair.
\( \textbf{Diagonal distance:} \)
\[
r_{\text{diag}}=a\sqrt{2}
\]
\( \textbf{Total diagonal contribution:} \)
\[
U_{\text{diag}}=2\left(\frac{kq^2}{a\sqrt{2}}\right)=\frac{\sqrt{2}kq^2}{a}
\]
\( \textbf{Total energy:} \)
\[
U=-\frac{4kq^2}{a}+\frac{\sqrt{2}kq^2}{a}
\]
\( \textbf{Final answer:} \) \(U=\frac{kq^2}{a}\left(-4+\sqrt{2}\right)\), with attractive side-pair contributions dominating.
314. In a one-dimensional region, the potential varies as \(V=80x-20x^2\), where \(V\) is in \(\text{V}\) and \(x\) is in \(\text{m}\). What is the electric field component at \(x=3\,\text{m}\)?
ⓐ. \(-20\,\text{V m}^{-1}\)
ⓑ. \(-40\,\text{V m}^{-1}\)
ⓒ. \(+20\,\text{V m}^{-1}\)
ⓓ. \(+40\,\text{V m}^{-1}\)
Correct Answer: \(+40\,\text{V m}^{-1}\)
Explanation: \( \textbf{Given potential:} \)
\[
V=80x-20x^2
\]
\( \textbf{Field-potential relation:} \)
\[
E_x=-\frac{dV}{dx}
\]
\( \textbf{Differentiate:} \)
\[
\frac{dV}{dx}=80-40x
\]
\( \textbf{At \(x=3\,\text{m}\):} \)
\[
\frac{dV}{dx}=80-40(3)=80-120=-40\,\text{V m}^{-1}
\]
\( \textbf{Apply negative sign:} \)
\[
E_x=-(-40)=+40\,\text{V m}^{-1}
\]
\( \textbf{Direction check:} \) At this point, potential decreases as \(x\) increases, so the electric field is along \(+x\).
\( \textbf{Final answer:} \) \(E_x=+40\,\text{V m}^{-1}\).
315. Use the arrangement described below and answer the question.
Three equipotential surfaces are marked \(S_1\), \(S_2\), and \(S_3\), with potentials \(120\,\text{V}\), \(80\,\text{V}\), and \(40\,\text{V}\), respectively. The distance between \(S_1\) and \(S_2\) is \(2.0\,\text{cm}\), while the distance between \(S_2\) and \(S_3\) is \(4.0\,\text{cm}\).
What is the ratio of electric field magnitudes in the two regions?
ⓐ. \(E_{12}:E_{23}=1:2\)
ⓑ. \(E_{12}:E_{23}=4:1\)
ⓒ. \(E_{12}:E_{23}=1:1\)
ⓓ. \(E_{12}:E_{23}=2:1\)
Correct Answer: \(E_{12}:E_{23}=2:1\)
Explanation: \( \textbf{Potential difference between \(S_1\) and \(S_2\):} \)
\[
|\Delta V_{12}|=120\,\text{V}-80\,\text{V}=40\,\text{V}
\]
\( \textbf{Potential difference between \(S_2\) and \(S_3\):} \)
\[
|\Delta V_{23}|=80\,\text{V}-40\,\text{V}=40\,\text{V}
\]
\( \textbf{Distances:} \)
\[
d_{12}=2.0\,\text{cm}=0.020\,\text{m}
\]
\[
d_{23}=4.0\,\text{cm}=0.040\,\text{m}
\]
\( \textbf{Field magnitude estimate:} \)
\[
E=\frac{|\Delta V|}{d}
\]
\( \textbf{First region:} \)
\[
E_{12}=\frac{40}{0.020}=2.0\times10^3\,\text{V m}^{-1}
\]
\( \textbf{Second region:} \)
\[
E_{23}=\frac{40}{0.040}=1.0\times10^3\,\text{V m}^{-1}
\]
\( \textbf{Final answer:} \) \(E_{12}:E_{23}=2:1\), because the same potential drop occurs over half the distance.
316. A conducting sphere of radius \(R\) carries charge \(Q\). A small charge \(q\) is moved slowly from the surface of the sphere to a point at distance \(2R\) from the centre. What is the work done by the external agent?
ⓐ. \(+\frac{kQq}{R}\)
ⓑ. \(-\frac{kQq}{2R}\)
ⓒ. \(0\)
ⓓ. \(+\frac{kQq}{2R}\)
Correct Answer: \(-\frac{kQq}{2R}\)
Explanation: \( \textbf{Potential of charged conducting sphere outside:} \) For \(r\ge R\),
\[
V=k\frac{Q}{r}
\]
\( \textbf{Initial point:} \) At the surface, \(r=R\), so
\[
V_i=k\frac{Q}{R}
\]
\( \textbf{Final point:} \) At \(r=2R\),
\[
V_f=k\frac{Q}{2R}
\]
\( \textbf{Potential difference:} \)
\[
\Delta V=V_f-V_i=\frac{kQ}{2R}-\frac{kQ}{R}
\]
\( \textbf{Simplification:} \)
\[
\Delta V=-\frac{kQ}{2R}
\]
\( \textbf{Slow external work:} \)
\[
W_{\text{ext}}=q\Delta V
\]
\( \textbf{Final result:} \)
\[
W_{\text{ext}}=-\frac{kQq}{2R}
\]
\( \textbf{Physical meaning:} \) If \(Qq\gt0\), the electric field helps the outward motion, so the external agent does negative work.
\( \textbf{Final answer:} \) \(W_{\text{ext}}=-\frac{kQq}{2R}\).
317. A parallel-plate capacitor is charged by a battery and then disconnected. Its plate separation is increased from \(d\) to \(3d\), and then a dielectric of constant \(K=3\) is fully inserted. What happens to the final voltage compared with the original voltage \(V_0\)?
ⓐ. It becomes \(3V_0\)
ⓑ. It becomes \(\frac{V_0}{3}\)
ⓒ. It becomes \(9V_0\)
ⓓ. It becomes \(V_0\)
Correct Answer: It becomes \(V_0\)
Explanation: \( \textbf{Initial capacitance:} \)
\[
C_0=\frac{\varepsilon_0A}{d}
\]
\( \textbf{Disconnected condition:} \) Charge remains fixed throughout the changes.
\( \textbf{Effect of increasing separation to \(3d\):} \)
\[
C_1=\frac{\varepsilon_0A}{3d}=\frac{C_0}{3}
\]
\( \textbf{Effect of fully inserting dielectric \(K=3\):} \)
\[
C_f=KC_1=3\left(\frac{C_0}{3}\right)=C_0
\]
\( \textbf{Voltage relation at fixed charge:} \)
\[
V=\frac{Q}{C}
\]
\( \textbf{Since \(C_f=C_0\):} \)
\[
V_f=\frac{Q}{C_0}=V_0
\]
\( \textbf{Final answer:} \) The final voltage becomes \(V_0\), because the dielectric increase exactly cancels the separation increase.
318. A capacitor is connected to a battery of voltage \(V\). Its plate separation is doubled while the battery remains connected. What happens to charge and energy stored?
ⓐ. \(Q\) doubles and \(U\) doubles
ⓑ. \(Q\) becomes half and \(U\) becomes one-fourth
ⓒ. \(Q\) becomes half and \(U\) becomes half
ⓓ. \(Q\) remains same and \(U\) doubles
Correct Answer: \(Q\) becomes half and \(U\) becomes half
Explanation: \( \textbf{Capacitance relation:} \)
\[
C=\frac{\varepsilon_0A}{d}
\]
\( \textbf{Doubling separation:} \) If \(d\) becomes \(2d\), capacitance becomes \(\frac{C}{2}\).
\( \textbf{Battery-connected condition:} \) The voltage \(V\) remains fixed.
\( \textbf{Charge relation:} \)
\[
Q=CV
\]
\( \textbf{Charge change:} \) Since \(C\) becomes half and \(V\) is fixed, \(Q\) becomes half.
\( \textbf{Energy relation at fixed voltage:} \)
\[
U=\frac{1}{2}CV^2
\]
\( \textbf{Energy change:} \) Since \(C\) becomes half and \(V\) is fixed, \(U\) becomes half.
\( \textbf{Final answer:} \) \(Q\) becomes half and \(U\) becomes half.
319. A dielectric slab of thickness \(t=\frac{3d}{4}\) and dielectric constant \(K=3\) is inserted into a parallel-plate capacitor of plate separation \(d\), covering the full plate area. What is the capacitance in terms of \(C_0=\frac{\varepsilon_0A}{d}\)?
ⓐ. \(2C_0\)
ⓑ. \(\frac{3C_0}{2}\)
ⓒ. \(\frac{C_0}{2}\)
ⓓ. \(\frac{4C_0}{3}\)
Correct Answer: \(2C_0\)
Explanation: \( \textbf{Partial slab formula:} \)
\[
C=\frac{\varepsilon_0A}{d-t+\frac{t}{K}}
\]
\( \textbf{Given values:} \)
\[
t=\frac{3d}{4}, \quad K=3
\]
\( \textbf{Effective separation:} \)
\[
d-\frac{3d}{4}+\frac{\frac{3d}{4}}{3}
\]
\( \textbf{Simplify terms:} \)
\[
d-\frac{3d}{4}=\frac{d}{4}
\]
\[
\frac{\frac{3d}{4}}{3}=\frac{d}{4}
\]
\( \textbf{Total effective separation:} \)
\[
\frac{d}{4}+\frac{d}{4}=\frac{d}{2}
\]
\( \textbf{New capacitance:} \)
\[
C=\frac{\varepsilon_0A}{d/2}=2\frac{\varepsilon_0A}{d}
\]
\( \textbf{Final answer:} \) \(C=2C_0\).
320. A \(2.0\,\mu\text{F}\) capacitor and a \(4.0\,\mu\text{F}\) capacitor are connected in series across \(18\,\text{V}\). A dielectric is then fully inserted only into the \(2.0\,\mu\text{F}\) capacitor, making its capacitance \(6.0\,\mu\text{F}\), while the battery remains connected. What is the new charge on each capacitor?
ⓐ. \(43.2\,\mu\text{C}\)
ⓑ. \(72.0\,\mu\text{C}\)
ⓒ. \(18.0\,\mu\text{C}\)
ⓓ. \(36.0\,\mu\text{C}\)
Correct Answer: \(43.2\,\mu\text{C}\)
Explanation: \( \textbf{After dielectric insertion:} \) The series capacitors are \(6.0\,\mu\text{F}\) and \(4.0\,\mu\text{F}\).
\( \textbf{Series equivalent capacitance:} \)
\[
C_{\text{eq}}=\frac{C_1C_2}{C_1+C_2}
\]
\( \textbf{Substitution:} \)
\[
C_{\text{eq}}=\frac{(6.0)(4.0)}{6.0+4.0}\,\mu\text{F}
\]
\( \textbf{Simplification:} \)
\[
C_{\text{eq}}=\frac{24.0}{10.0}\,\mu\text{F}=2.4\,\mu\text{F}
\]
\( \textbf{Battery voltage:} \) The total voltage remains \(18\,\text{V}\).
\( \textbf{Series charge:} \)
\[
Q=C_{\text{eq}}V
\]
\( \textbf{Calculation:} \)
\[
Q=(2.4\,\mu\text{F})(18\,\text{V})=43.2\,\mu\text{C}
\]
\( \textbf{Series condition:} \) Each capacitor carries the same charge magnitude.
\( \textbf{Final answer:} \) The new charge on each capacitor is \(43.2\,\mu\text{C}\).
