Capacitors

Answer: (a)

1. Capacitance is constant: $C = \frac{Q}{V} = \frac{Q_1}{V - V_1}$.
2. $Q(V - V_1) = Q_1 V \Rightarrow V(Q - Q_1) = QV_1$
3. So the answer is (a) $V = \frac{QV_1}{Q - Q_1}$.

Answer: (d)

1. Capacitance of a sphere: $C = 4\pi\varepsilon_0 R$.
2. $\frac{V}{A} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3} \Rightarrow R = \frac{3V}{A}$
3. $C = 4\pi\varepsilon_0 \cdot \frac{3V}{A}$
4. So the answer is (d) $12\pi\varepsilon_0\frac{V}{A}$.

Answer: (d)

1. Capacitance of a sphere: $C = 4\pi\varepsilon_0 R$.
2. $\frac{V}{A} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3} \Rightarrow R = \frac{3V}{A}$
3. $C = 4\pi\varepsilon_0 \cdot \frac{3V}{A}$
4. So the answer is (d) $12\pi\epsilon_0 V/A$.

Answer: (a)

1. Capacitance is constant: $C = \frac{Q}{V} = \frac{Q_1}{V - V_1}$.
2. $Q(V - V_1) = Q_1 V \Rightarrow V(Q - Q_1) = QV_1$
3. So the answer is (a) $V = \frac{QV_1}{Q - Q_1}$.

Answer: (b)

1. For a capacitor $V = \frac{Q}{C}$, so the slope of the $V$–$Q$ line is $\frac{1}{C}$.
2. For the same $Q$, $V_A < V_B$, so line A has the smaller slope.
3. Smaller slope means larger capacitance: $C_A > C_B$.
4. So the answer is (b) capacity of $A$ is higher than $B$.

Answer: (a)

1. For a capacitor $V = \frac{Q}{C}$, so the slope of the $V$–$Q$ line is $\frac{1}{C}$.
2. For the same $Q$, $V_A < V_B$, so line A has the smaller slope.
3. Smaller slope means larger capacitance: $C_A > C_B$.
4. So the answer is (a) capacitor A has the higher capacity.

Answer: (b)

1. $C = \frac{q}{V}$, but $q$ and $V$ change in the same proportion, so their ratio is fixed.
2. $C$ depends only on size, shape, separation of conductors and the medium, e.g. $C = \frac{\varepsilon_0 A}{d}$.
3. So the answer is (b) the geometry of the capacitor.

Answer: (b)

1. Potential difference $V = 20 - (-20) = 40\text{ V}$.
2. $C = \frac{Q}{V} = \frac{40}{40}$
3. So the answer is (b) $1\text{ F}$.

Answer: (c)

1. Charge on a conductor stays on its outer surface, so a solid and a hollow sphere of the same radius have the same $C = 4\pi\varepsilon_0 R$.
2. At the same potential they hold the same charge $Q = CV$, so A is false.
3. R is true: $C \propto R$.
4. So the answer is (c) A is false but R is true.

Answer: (c)

1. Same potential: $Q = CV$, and $C \propto r$, so $Q \propto r$.
2. $\sigma = \frac{Q}{4\pi r^2} \propto \frac{r}{r^2} = \frac{1}{r}$
3. $\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1} = \frac{2}{3}$
4. So the answer is (c) $\frac{2}{3}$.

Answer: (a)

1. $V = \frac{Q}{C}$ with constant $C$, so $V \propto Q$: a straight line through the origin.
2. At $Q = 5\text{ C}$: $V = \frac{5}{2\times10^{-6}} = 2.5\times10^{6}\text{ V}$.
3. So the answer is (a), the straight line ending at $2.5\times10^{6}\text{ V}$.

Answer: (a)

1. Isolated sphere: $C = 4\pi\varepsilon_0 R_1$.
2. With earthed outer shell (spherical capacitor): $C' = \frac{4\pi\varepsilon_0 R_1R_2}{R_2 - R_1}$.
3. $C' = nC \Rightarrow \frac{R_2}{R_2 - R_1} = n \Rightarrow R_2(n-1) = nR_1$
4. So the answer is (a) $\frac{R_2}{R_1} = \frac{n}{n-1}$.

Answer: (d)

1. $U = \frac{Q^2}{2C} = \frac{Q^2}{8\pi\varepsilon_0 R} \Rightarrow R = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Q^2}{2U}$
2. $R = \frac{9\times10^{9}\times(4\times10^{-6})^2}{2\times4.5} = \frac{0.144}{9}$
3. $R = 0.016\text{ m} = 16\text{ mm}$
4. So the answer is (d) 16 mm.

Answer: (a)

1. $C = 4\pi\varepsilon_0 R = \frac{R}{9\times10^{9}}$
2. $C = \frac{1}{9\times10^{9}} \approx 1.1\times10^{-10}\text{ F}$
3. So the answer is (a) $1.1\times10^{-10}$.

Answer: (c)

1. $C = \frac{\varepsilon_0 \pi r^2}{d} \propto \frac{r^2}{d}$
2. $\frac{C_2}{C_1} = \frac{(\sqrt{3})^2}{1/2} = 6$
3. So the answer is (c) $C_1 : C_2 = 1 : 6$.

Answer: (c)

1. Charge on outer faces $= \frac{q_1 + q_2}{2}$ each; inner faces carry $\pm\frac{q_1 - q_2}{2}$.
2. Only the inner charges make the field between the plates, so effective charge $Q = \frac{q_1 - q_2}{2}$.
3. $V = \frac{Q}{C} = \frac{q_1 - q_2}{2C}$
4. So the answer is (c) $\frac{q_1 - q_2}{2C}$.

Answer: (d)

1. $C = \frac{\varepsilon_0 lb}{d} \propto \frac{lb}{d}$. Original: $\frac{3\times1}{3} = 1$ (cm² per μm); need 10.
2. A: $\frac{30}{1} = 30$; B: $\frac{3}{30} = 0.1$; C: $\frac{30}{3} = 10$ ✓
3. D: $\frac{1}{10} = 0.1$; E: $\frac{10}{1} = 10$ ✓
4. So the answer is (d) C and E only.

Answer: (c)

1. Charge on outer faces $= \frac{q_1 + q_2}{2}$ each; inner faces carry $\pm\frac{q_1 - q_2}{2}$.
2. Only the inner charges make the field between the plates, so effective charge $Q = \frac{q_1 - q_2}{2}$.
3. $V = \frac{Q}{C} = \frac{q_1 - q_2}{2C}$
4. So the answer is (c) $\frac{q_1 - q_2}{2C}$.

Answer: (b)

1. Acceleration: $a = \frac{eE}{m} = \frac{1.6\times10^{-19}\times910}{9.1\times10^{-31}} = 1.6\times10^{14}\text{ m/s}^2$ ($9.1\text{ V/cm} = 910\text{ V/m}$).
2. Time between plates: $t = \frac{0.1}{10^{6}} = 10^{-7}\text{ s}$
3. $v_y = at = 1.6\times10^{14}\times10^{-7} = 1.6\times10^{7}\text{ m/s}$
4. So the answer is (b) $16\times10^{6}\text{ m/s}$.

Answer: (c)

1. $C = \frac{\varepsilon_0 \pi r^2}{d} \propto \frac{r^2}{d}$
2. $\frac{C_2}{C_1} = \frac{(\sqrt{2})^2}{1/2} = 4$
3. So the answer is (c) $C_1 : C_2 = 1 : 4$.

Answer: (b)

1. Uniform field: $E = \frac{V}{t}$, and $V = \frac{Q}{C}$.
2. So the answer is (b) $E = \frac{Q}{Ct}$.

Answer: (b)

1. Gauss's law: $\phi = \frac{q_{enc}}{\varepsilon_0}$.
2. The surface encloses only the positive plate, charge $Q = CV$.
3. So the answer is (b) $\frac{CV}{\varepsilon_0}$.

Answer: (a)

1. $C = \frac{\varepsilon_0 A}{d}$
2. $C' = \frac{\varepsilon_0 (A/2)}{2d} = \frac{C}{4}$
3. So the answer is (a) $\frac{C}{4}$.

Answer: (c)

1. Charge on outer faces $= \frac{q_1 + q_2}{2}$ each; inner faces carry $\pm\frac{q_1 - q_2}{2}$.
2. Only the inner charges make the field between the plates, so effective charge $Q = \frac{q_1 - q_2}{2}$.
3. $V = \frac{Q}{C} = \frac{q_1 - q_2}{2C}$
4. So the answer is (c) $\frac{q_1 - q_2}{2C}$.

Answer: (a)

1. Between the plates each plate gives $\frac{\sigma}{2\varepsilon_0}$; together $E = \frac{\sigma}{\varepsilon_0}$.
2. With one plate removed, $E' = \frac{\sigma}{2\varepsilon_0} = \frac{E}{2}$.
3. $F' = \frac{F}{2} = \frac{10}{2}$
4. So the answer is (a) $5\text{ N}$.

Answer: (c)

1. $C = \frac{\varepsilon_0 A}{d} \propto \frac{1}{d}$
2. $d \to \frac{d}{2} \Rightarrow C$ doubles.
3. So the answer is (c) $C_2 = 2C_1$.

Answer: (c)

1. Slab region and air region are in series. Take $C_2 = \frac{K\varepsilon_0 A}{t}$ (slab, beside the $+V_2$ plate) and $C_1 = \frac{\varepsilon_0 A}{d-t}$ (air).
2. Total p.d. $= V_2 - (-V_1) = V_1 + V_2$, so $Q = \frac{C_1C_2}{C_1 + C_2}(V_1 + V_2)$.
3. Field in air: $E = \frac{Q/C_1}{d - t} = \frac{C_2(V_1 + V_2)}{(C_1 + C_2)(d - t)}$
4. Equilibrium: $\tan\theta = \frac{qE}{mg}$
5. So the answer is (c) $\tan^{-1}\left[\frac{q}{mg}\times\frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}\right]$.

Answer: (a)

1. The field is perpendicular to the plates, so the velocity component parallel to the plates is unchanged.
2. $v_1\cos\alpha = v_2\cos\beta \Rightarrow \frac{v_1}{v_2} = \frac{\cos\beta}{\cos\alpha}$
3. $\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2}$
4. So the answer is (a) $\frac{\cos^2\beta}{\cos^2\alpha}$.

Answer: (c)

1. At distance $x$ from the narrow edge the gap is $d + \alpha x$; strips of width $dx$ are in parallel.
2. $C = \int_0^a \frac{\varepsilon_0 a\,dx}{d + \alpha x} = \frac{\varepsilon_0 a}{\alpha}\ln\left(1 + \frac{\alpha a}{d}\right)$
3. Using $\ln(1+x) \approx x - \frac{x^2}{2}$: $C \approx \frac{\varepsilon_0 a}{\alpha}\left(\frac{\alpha a}{d} - \frac{\alpha^2a^2}{2d^2}\right)$
4. So the answer is (c) $\frac{\varepsilon_0 a^2}{d}\left(1 - \frac{\alpha a}{2d}\right)$.

Answer: (b)

1. $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \Rightarrow Q = \varepsilon_0 AE$
2. $Q = 8.85\times10^{-12}\times1\times100 = 8.85\times10^{-10}\text{ C}$
3. So the answer is (b) $8.85\times10^{-10}\text{ C}$.

Answer: (a)

1. With plate charges $q_1$, $q_2$, the inner faces carry $\pm\frac{q_1 - q_2}{2}$.
2. $V = \frac{q_1 - q_2}{2C} = \frac{(4 - 2)\ \mu\text{C}}{2\times1\ \mu\text{F}}$
3. So the answer is (a) $1\text{ V}$.

Answer: (c)

1. On the bob: weight $mg$ downward and $qE$ horizontal, perpendicular to each other.
2. Effective gravity: $g_{eff} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$
3. $T = 2\pi\sqrt{\frac{L}{g_{eff}}}$
4. So the answer is (c) $2\pi\sqrt{\frac{L}{\sqrt{g^2 + \frac{q^2E^2}{m^2}}}}$.

Answer: (c)

1. Energy density $u = \frac{1}{2}\varepsilon_0 E^2$.
2. Volume between plates $= Ad$, so $U = \frac{1}{2}\varepsilon_0 E^2 Ad$.
3. So the answer is (c) $\frac{1}{2}A\epsilon_0 E^2 d$.

Answer: (a)

1. $W = \frac{Q^2}{2C} \propto Q^2$ for fixed $C$.
2. $Q \to 3Q \Rightarrow W' = 9W$
3. So the answer is (a) $9W$.

Answer: (c)

1. Each plate feels the field of the other plate only: $E' = \frac{\sigma}{2\varepsilon_0}$.
2. $F = QE' = \frac{Q^2}{2\varepsilon_0 A}$
3. For an isolated capacitor $Q$ is fixed, and $d$ does not appear in $F$.
4. So the answer is (c) independent of $d$.

Answer: (c)

1. Total field between plates $E = \frac{\sigma}{\varepsilon_0}$; half of it, $\frac{E}{2}$, is due to each plate.
2. A plate cannot push itself, so it feels only the other plate's field $\frac{E}{2}$.
3. $F = Q \cdot \frac{E}{2}$
4. So the answer is (c) $\frac{QE}{2}$.

Answer: (a)

1. $U = \frac{1}{2}CV^2 \Rightarrow \frac{dU}{dt} = CV\frac{dV}{dt}$
2. $\frac{dU}{dt} = 5 \times 10^{-6} \times 4 \times 0.6$
3. $\frac{dU}{dt} = 12 \times 10^{-6}\text{ J/s}$
4. So the answer is (a) $12 \times 10^{-6}$ J/s.

Answer: (a)

1. Battery connected: $V$ constant, $U = \frac{1}{2}CV^2 = \frac{\varepsilon_0 AV^2}{2x}$.
2. $\frac{dU}{dt} = -\frac{\varepsilon_0 AV^2}{2x^2}\frac{dx}{dt} = -\frac{\varepsilon_0 AV^2 v}{2x^2}$
3. With $v$ constant, $\frac{dU}{dt} \propto x^{-2}$.
4. So the answer is (a) $\alpha = -2$.

Answer: (d)

1. $U = \frac{1}{2}CV^2$ with $C = \frac{\varepsilon_0 A}{d}$ and $V = Ed$.
2. $U = \frac{1}{2}\cdot\frac{\varepsilon_0 A}{d}\cdot E^2 d^2 = \frac{1}{2}\varepsilon_0 E^2 Ad$
3. So the answer is (d) $\frac{1}{2}\varepsilon_0 E^2 Ad$.

Answer: (c)

1. $U = \frac{q^2}{2C} \Rightarrow U \propto q^2$.
2. $\frac{(q+3)^2}{q^2} = 1.21 \Rightarrow \frac{q+3}{q} = 1.1$
3. $0.1q = 3 \Rightarrow q = 30\text{ C}$
4. So the answer is (c) 30 C.

Answer: (a)

1. $E = \frac{V}{d} = \frac{20}{10^{-6}} = 2 \times 10^{7}\text{ V/m}$
2. $u = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14}$
3. $u = 1770 \approx 1.8 \times 10^{3}\text{ J/m}^3$
4. So the answer is (a) $1.8 \times 10^{3}$ J/m$^3$.