Solution: $E_x = -\partial V/\partial x = -(6 - 8y^2)$. At $(1,1,1)$, $E_x = -(6 - 8) = 2$. $E_y = -\partial V/\partial y = -(-16xy - 8 + 6z)$. At $(1,1,1)$, $E_y = -(-16 - 8 + 6) = 18$. $E_z = -\partial V/\partial z = -(6y) = -6$. $|\vec{E}| = \sqrt{2^2 + 18^2 + (-6)^2} = \sqrt{4 + 324 + 36} = \sqrt{364} \approx 19.08 \text{ V/m}$.
Solution: $V_{inner} = 0 \Rightarrow \frac{kq}{R} + \frac{k(2Q)}{2R} = 0 \Rightarrow \frac{q}{R} + \frac{Q}{R} = 0 \Rightarrow q = -Q$.
Solution: $V_{outer} = 0 \Rightarrow \frac{kQ}{2R} + \frac{kq'}{2R} = 0 \Rightarrow q' = -Q$.
Solution: $dV = \frac{k(dq)}{r} = \frac{k(\rho \cdot 4\pi r^2 dr)}{r} = 4\pi k \rho r dr = \frac{1}{\epsilon_0} \rho_0 (1 - r/R) r dr$. Integrating from $0$ to $R$: $V = \frac{\rho_0}{\epsilon_0} \int_{0}^{R} (r - \frac{r^2}{R}) dr = \frac{\rho_0}{\epsilon_0} [\frac{R^2}{2} - \frac{R^3}{3R}] = \frac{\rho_0 R^2}{6\epsilon_0}$.
Solution: $\vec{E} = E\cos 30^\circ \hat{i} + E\sin 30^\circ \hat{j} = E(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j})$. $\vec{r}_{BA} = \vec{r}_A - \vec{r}_B = 2\hat{i} - 4\hat{j}$. $V_B - V_A = \vec{E} \cdot \vec{r}_{BA} = E(\sqrt{3} - 2)$. Thus $V_A - V_B = E(2 - \sqrt{3})$.
Solution: $V \propto \vec{p} \cdot \hat{r} = p \cos\theta$. For $V=0$, $\cos\theta = 0 \Rightarrow \theta = \pi/2$ or $3\pi/2$. This represents the y-axis ($x=0$). Locus is the y-axis.
Solution: The uncharged sphere will have induced charges, but the net induced charge is zero. The potential of the sphere is uniform. $V_{center} = V_{sphere} = V_{due\_to\_q} + V_{induced}$. Since all induced charges are at distance $R$ from center and sum to zero, $V_{induced} = k(0)/R = 0$. Thus, $V = kq/r$.
Solution: This is a linear quadrupole (two opposing dipoles). $V = \frac{k}{r}[q(1-a\cos\theta/r)^{-1} - 2q + q(1+a\cos\theta/r)^{-1}] \approx \frac{kq a^2}{r^3} (3\cos^2\theta - 1)$. It falls off as $1/r^3$. It is an electric quadrupole.
Solution: Let point be origin. Wire from $x=d$ to $x=d+L$. $dV = \frac{k(\lambda dx)}{x}$. $V = k\lambda \int_{d}^{d+L} \frac{dx}{x} = k\lambda \ln(\frac{d+L}{d}) = \frac{\lambda}{4\pi\epsilon_0} \ln(1 + \frac{L}{d})$.
Solution: $V_1 = \frac{kQ_1}{R} + \frac{kQ_2}{\sqrt{R^2+R^2}} = \frac{kQ_1}{R} + \frac{kQ_2}{R\sqrt{2}}$. $V_2 = \frac{kQ_2}{R} + \frac{kQ_1}{R\sqrt{2}}$. Work $W = q(V_2 - V_1) = \frac{kq}{R} [(Q_2 - Q_1) - \frac{Q_2 - Q_1}{\sqrt{2}}] = \frac{kq(Q_2 - Q_1)}{R} (1 - \frac{1}{\sqrt{2}})$.
Solution: $V_A - V_O = -\int_{0}^{2} E_x dx = -\int_{0}^{2} 30x^2 dx = -30 [x^3/3]_0^2 = -10 (8) = -80 \text{ V}$.
Solution: $V = \frac{k}{r}(q - q + q - q) = 0$. For $\vec{E}$, diagonals cancel each other out identically since opposite charges are same sign? Wait: A(q), B(-q), C(q), D(-q). Diagonals are AC (q, q) and BD (-q, -q). E-fields from A and C cancel. E-fields from B and D cancel. Thus $\vec{E} = 0$.
Solution: $E_{axis} = \frac{kQx}{(R^2+x^2)^{3/2}} \approx \frac{kQx}{R^3}$ for $x \ll R$. Restoring force $F = -qE = -\frac{kqQ}{R^3}x$. Thus $a = -\omega^2 x$ where $\omega = \sqrt{\frac{kqQ}{mR^3}}$. $T = 2\pi \sqrt{\frac{mR^3}{kqQ}} = 2\pi \sqrt{\frac{4\pi\epsilon_0 m R^3}{qQ}}$.
Solution: Between sheets, $E = \frac{\sigma_1}{2\epsilon_0} - \frac{-\sigma_2}{2\epsilon_0}$ if opposite, but problem says $\sigma_1, \sigma_2$. If both positive, $E = \frac{\sigma_1 - \sigma_2}{2\epsilon_0}$. $\Delta V = Ed = \frac{(\sigma_1 - \sigma_2)d}{2\epsilon_0}$.
Solution: $Q = Ze = 79 \times 1.6 \times 10^{-19} \text{ C}$. $V = \frac{kQ}{R} = \frac{9 \times 10^9 \times 79 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-15}} = \frac{1137.6 \times 10^{-10}}{6.6 \times 10^{-15}} \approx 1.72 \times 10^7 \text{ V} = 17.2 \text{ MV}$.
Solution: By superposition, $E_{shell} = E_{rem} + E_{hole} = 0$ just inside, and $\sigma/\epsilon_0$ just outside. At hole surface, $E = \sigma / 2\epsilon_0 = \frac{Q}{2\epsilon_0(4\pi R^2)} = \frac{Q}{8\pi\epsilon_0 R^2}$. Potential $V \approx \frac{Q}{4\pi\epsilon_0 R}$ (hole effect is negligible for $V$ if $r \ll R$).
Solution: The surface is a sphere. E-field lines are strictly perpendicular to equipotential surfaces. Therefore, the E-field lines are purely radial straight lines passing through the origin.
Solution: $\vec{E} = -\frac{dV}{dr} \hat{r} = -(-\frac{A}{r^2} - \frac{2B}{r^3}) \hat{r} = (\frac{A}{r^2} + \frac{2B}{r^3}) \hat{r}$.
Solution: $\Delta V$ depends only on the charge of the inner sphere $Q$, because the charge on the outer shell contributes equally to the potential of both the inner sphere and the outer shell. Since inner charge $Q$ is unchanged, $\Delta V$ remains $V$.
Solution: Field of $q$ is $E = \frac{kq}{x^2}$. Force on dipole $F = p_0 \frac{dE}{dx} = p_0 (-\frac{2kq}{x^3})$. Magnitude is $\frac{2kqp_0}{x^3}$, direction is towards origin (attractive).
Solution: Largest charges should be farthest apart. Place $8q$ at $x=0$, $2q$ at $x=9$. Let $q$ be at $x$. $U = k[ \frac{8q \cdot 2q}{9} + \frac{8q \cdot q}{x} + \frac{2q \cdot q}{9-x} ]$. For min $U$, $\frac{dU}{dx} = 0 \Rightarrow -\frac{8}{x^2} + \frac{2}{(9-x)^2} = 0 \Rightarrow \frac{4}{x^2} = \frac{1}{(9-x)^2} \Rightarrow \frac{2}{x} = \frac{1}{9-x} \Rightarrow 18 - 2x = x \Rightarrow 3x = 18 \Rightarrow x = 6 \text{ cm}$.
Solution: $U = \int_0^R V(r) dq$. Integrating assembling shells gives $U = \frac{3}{5} \frac{kQ^2}{R} = \frac{3Q^2}{20\pi\epsilon_0 R}$.
Solution: $U = k[\frac{qQ_1}{r} + \frac{qQ_2}{2r} + \frac{Q_1Q_2}{r}] = 0$. $\Rightarrow qQ_1 + \frac{1}{2}qQ_2 + Q_1Q_2 = 0$. This needs $q$ relation, wait, $Q_1, Q_2$ are independent? Usually this implies a specific ratio or $q$ is given. If question means $U$ independent of $q$, then $Q_1 + Q_2/2 = 0 \Rightarrow Q_1/Q_2 = -1/2$.
Solution: $K_i = U_f \Rightarrow 10 \times 10^6 \times 1.6 \times 10^{-19} = \frac{k(2e)(50e)}{r_0}$. $1.6 \times 10^{-12} = \frac{9 \times 10^9 \times 100 \times (1.6 \times 10^{-19})^2}{r_0} \Rightarrow r_0 = \frac{9 \times 10^{11} \times 2.56 \times 10^{-38}}{1.6 \times 10^{-12}} = 1.44 \times 10^{-14} \text{ m} = 14.4 \text{ fm}$.
Solution: Energy conservation: $2(\frac{1}{2}mv^2) = \frac{kq^2}{r_{min}} \Rightarrow mv^2 = \frac{kq^2}{r_{min}} \Rightarrow r_{min} = \frac{kq^2}{mv^2} = \frac{q^2}{4\pi\epsilon_0 mv^2}$.
Solution: Work done by electric field $W = Fx = (qE)x$. By Work-Energy Theorem, $K = W = qEx$.
Solution: $F = p \frac{dE}{dx} = p \frac{d}{dx}(\alpha x) = p\alpha$.
Solution: Field of $p_1$ at position of $p_2$ is $E = \frac{2kp_1}{r^3}$. $U = -\vec{p}_2 \cdot \vec{E} = -p_2 (\frac{2kp_1}{r^3}) = -\frac{1}{4\pi\epsilon_0} \frac{2p_1 p_2}{r^3}$.
Solution: $\theta_1 = 0^\circ$, $\theta_2 = 180^\circ$. $W = U_f - U_i = (-pE\cos 180^\circ) - (-pE\cos 0^\circ) = (+pE) - (-pE) = 2pE$.
Solution: Inside sphere, $E = \frac{kQr}{R^3}$. Energy density $u = \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 \left( \frac{Qr}{4\pi\epsilon_0 R^3} \right)^2 = \frac{Q^2 r^2}{32\pi^2 \epsilon_0 R^6}$.
Solution: Number of pairs $n = \frac{4(4-1)}{2} = 6$. All pairs are at distance $a$. $W = U = 6 \times \frac{kq^2}{a} = \frac{6q^2}{4\pi\epsilon_0 a}$.
Solution: $K = qV = 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$. $p = \sqrt{2mK} = \sqrt{2(1.67\times 10^{-27})(1.6\times 10^{-13})} \approx 2.3 \times 10^{-20} \text{ kg m/s}$. $\lambda = h/p = 6.63\times 10^{-34} / 2.3\times 10^{-20} \approx 2.88 \times 10^{-14} \text{ m}$.
Solution: $mg = qE \Rightarrow m(10) = (3\times 10^{-6})(10^5) = 0.3 \Rightarrow m = 0.03 \text{ kg} = 30 \text{ grams}$.
Solution: $T\sin\theta = F_e = \frac{kq^2}{(2l\sin\theta)^2}$. $T\cos\theta = mg$. Dividing gives $\tan\theta = \frac{kq^2}{4l^2\sin^2\theta \cdot mg} \Rightarrow q = \sqrt{\frac{4l^2 mg \sin^2\theta \tan\theta}{k}} = 4l\sin\theta\sqrt{\pi\epsilon_0 mg\tan\theta}$.
Solution: Potential at center of a cubic shell of charge $Q$ is difficult to compute directly, but by dimensional analysis and integrating $\sigma$, $V_0 = \text{constant} \times \frac{kQ}{a}$. Interaction energy is $qV_0$.
Solution: $\vec{E}$ is radial. $\vec{p}$ is parallel to $\vec{E}$. $U = -pE = -p(\frac{kq}{r^2})$. Force $F = p\frac{dE}{dr} = -2\frac{kqp}{r^3}$ (attractive). Displacing radially, force increases as $r$ decreases, pulling it in. Hence, unstable equilibrium in radial direction.
Solution: $U_i = U_{self} + U_{interaction} = \frac{kq^2}{2R_1} + \frac{kqq_0}{R_1}$. $U_f = \frac{kq^2}{2R_2} + \frac{kqq_0}{R_2}$. Work done by external agent $W = U_f - U_i = (\frac{kq^2}{2} + kqq_0) (\frac{1}{R_2} - \frac{1}{R_1})$. Since $R_2 > R_1$, $W < 0$.
Solution: The potential at the center of the base of a solid cone is $V = \frac{\rho}{4\epsilon_0} [\sqrt{R^2+h^2} - h]h$? Wait, precise integration gives $V = \frac{\rho}{2\epsilon_0} [\sqrt{R^2+h^2} - h]h$? Advanced integration needed. Let's provide formula: $W = q V_{cone}$.
Solution: $U = U_{self1} + U_{self2} + U_{interaction} = \frac{kq_1^2}{2r_1} + \frac{kq_2^2}{2r_2} + \frac{kq_1q_2}{r_2}$.
Solution: $\frac{1}{2}mv_0^2 = \frac{kQq}{r_{min}} \Rightarrow r_{min} \propto 1/v_0^2$. If $v_0$ is doubled, $r_{min}$ becomes one-fourth ($1/4$) of its original value.
Solution: Treat as series combination of $dC = \frac{K(x)\epsilon_0 A}{dx}$. $1/C = \int_{0}^{d} \frac{dx}{(K_0+\alpha x)\epsilon_0 A} = \frac{1}{\alpha \epsilon_0 A} [\ln(K_0+\alpha x)]_0^d = \frac{1}{\alpha \epsilon_0 A} \ln(1 + \frac{\alpha d}{K_0})$. $C = \frac{\alpha \epsilon_0 A}{\ln(1 + \alpha d/K_0)}$.
Solution: $C_{eq} = \left(\frac{1}{C_1} + \frac{1}{C_2 + C_{eq}}\right)^{-1} \Rightarrow \frac{1}{C_{eq}} = \frac{C_1 + C_2 + C_{eq}}{C_1(C_2+C_{eq})} \Rightarrow C_{eq}^2 + C_2 C_{eq} - C_1 C_2 = 0$. Solving quadratic: $C_{eq} = \frac{-C_2 + \sqrt{C_2^2 + 4C_1 C_2}}{2}$.
Solution: This is a parallel combination of two capacitors: $C_{air}$ (length $a-x$) and $C_{dielectric}$ (length $x$). $C = C_{air} + C_{diel} = \frac{\epsilon_0 a(a-x)}{d} + \frac{K\epsilon_0 ax}{d} = \frac{\epsilon_0 a}{d}[a + (K-1)x]$.
Solution: $F = +\frac{d U}{d x}$ (at constant V) or $\frac{1}{2} V^2 \frac{dC}{dx}$. $\frac{dC}{dx} = \frac{\epsilon_0 a(K-1)}{d}$. $F = \frac{\epsilon_0 a(K-1)V^2}{2d}$ (pulling inwards).
Solution: At constant Q, $F = -\frac{dU}{dx} = -\frac{d}{dx}(\frac{Q^2}{2C}) = \frac{Q^2}{2C^2}\frac{dC}{dx}$. $F = \frac{Q^2 \epsilon_0 a(K-1)}{2d [\frac{\epsilon_0 a}{d}(a+(K-1)x)]^2} = \frac{Q^2 d (K-1)}{2\epsilon_0 a [a + (K-1)x]^2}$.
Solution: By symmetry, using nodes of same potential, $C_{eq} = \frac{6}{5}C$.
Solution: By symmetry, equivalent capacitance is $C_{eq} = \frac{4}{3}C$.
Solution: (a) Series: $C_a = \frac{\epsilon_0 A}{d/2K_1 + d/2K_2} = \frac{2\epsilon_0 A}{d} \frac{K_1 K_2}{K_1+K_2}$. (b) Parallel: $C_b = \frac{K_1\epsilon_0 (A/2)}{d} + \frac{K_2\epsilon_0 (A/2)}{d} = \frac{\epsilon_0 A}{2d} (K_1+K_2)$. Ratio $C_a / C_b = \frac{4 K_1 K_2}{(K_1+K_2)^2}$.
Solution: Consider shell $dr$. $dC = \frac{K_0/r \cdot \epsilon_0 4\pi r^2}{dr}$? No, series combination $dV = \frac{Qdr}{4\pi\epsilon_0 K_0/r \cdot r^2} = \frac{Qdr}{4\pi\epsilon_0 K_0 r}$. $V = \frac{Q}{4\pi\epsilon_0 K_0} \ln(b/a)$. $C = Q/V = \frac{4\pi\epsilon_0 K_0}{\ln(b/a)}$.
Solution: B is one terminal, AC is another. This is two capacitors in parallel: $C_{AB}$ and $C_{BC}$. $C_{AB} = \frac{\epsilon_0 A}{d}$. $C_{BC} = \frac{\epsilon_0 A}{2d}$. Total $C_{eq} = \frac{\epsilon_0 A}{d} + \frac{\epsilon_0 A}{2d} = \frac{3\epsilon_0 A}{2d}$.
Solution: $Q_1 = 200\mu\text{C}, Q_2 = 200\mu\text{C}$. Opposite polarity means total charge $Q = 200 - 200 = 0$. Common potential $V=0$. Final energy $U_f = 0$. Initial $U_i = \frac{1}{2}(2)(100)^2 + \frac{1}{2}(4)(50)^2 = 10000 + 5000 = 15000 \mu\text{J} = 0.015 \text{ J}$. Heat = $0.015 \text{ J}$.
Solution: Metal $K=\infty$. $C = \frac{\epsilon_0 A}{d - t(1 - 1/\infty)} = \frac{\epsilon_0 A}{d-t}$.
Solution: Breakdown Field $E_{max} = V_b/d$. Max energy density $u = \frac{1}{2} K \epsilon_0 E_{max}^2$. Total max energy $U_{max} = u \times V_{ol} = \frac{1}{2} K \epsilon_0 (V_b/d)^2 V_{ol}$. Wait, usually $V_b$ is defined as voltage, or breakdown field $E_b$. If $V_b$ is breakdown voltage, $U_{max} = \frac{1}{2} C V_b^2$.
Solution: Energy conservation: Max electrical energy = Max magnetic energy. $\frac{1}{2} C V_0^2 = \frac{1}{2} L I_{max}^2 \Rightarrow I_{max} = V_0 \sqrt{C/L}$.
Solution: By symmetry and superposition principles used in infinite resistive grids, the equivalent capacitance for adjacent nodes in an infinite 2D square lattice of capacitors $C$ is $C_{eq} = 2C$. (Equivalent to resistor case $R/2$, so $1/C_{eq} = (1/C)/2 \Rightarrow C_{eq} = 2C$).
Solution: Volume $\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \Rightarrow R = 2r \Rightarrow r = R/2$. Capacitance $c = 4\pi\epsilon_0 r = 4\pi\epsilon_0 (R/2) = C/2 = 0.5 \mu\text{F}$.
Solution: Overlapping area $A(\theta) = \frac{1}{2}R^2 (\pi - \theta)$. Number of capacitors is $(N-1)$. $C(\theta) = \frac{(N-1)\epsilon_0 \frac{1}{2}R^2(\pi-\theta)}{d}$.
Solution: $U_i = \frac{1}{2}C V^2$. Battery connected $\Rightarrow V$ is constant. $C' = KC$. $U_f = \frac{1}{2}(KC)V^2 = K \cdot U_i$. Ratio $U_i / U_f = 1/K$.
Solution: Initial charge $Q = C_0 V$. Final charge $Q' = KC_0 V$. Extra charge supplied $\Delta Q = (K-1)C_0 V$. Work by battery $W = \Delta Q \cdot V = (K-1)C_0 V^2$.
Solution: Area is divided. Two capacitors in parallel, each with area $A/2$. $C_1 = \frac{K_1 \epsilon_0 (A/2)}{d} = K_1 (C_0/2) = 4(4) = 16\mu\text{F}$. $C_2 = K_2 (C_0/2) = 2(4) = 8\mu\text{F}$. Total $C = C_1 + C_2 = 16 + 8 = 24\mu\text{F}$.
