Solution: Electric potential is the work done by an external force in bringing a unit positive charge from infinity to that point. Inside a hollow charged spherical conductor, the electric field $E = 0$. Since $E = -\frac{dV}{dr}$, if $E=0$, then $dV=0$, which implies $V$ is a constant.
Solution: Let $P$ be at distance $x$ from positive charge. Internal point: $\frac{k(5 \times 10^{-8})}{x} + \frac{k(-3 \times 10^{-8})}{16-x} = 0 \Rightarrow \frac{5}{x} = \frac{3}{16-x} \Rightarrow 80 - 5x = 3x \Rightarrow 8x = 80 \Rightarrow x = 10 \text{ cm}$. External point (beyond negative charge): $\frac{5}{x} = \frac{3}{x-16} \Rightarrow 5x - 80 = 3x \Rightarrow 2x = 80 \Rightarrow x = 40 \text{ cm}$.
Solution: $R = 1.2 \text{ m}$. Total charge $Q = \sigma \times 4\pi R^2 = 80 \times 10^{-6} \times 4\pi \times (1.2)^2 = 1.45 \times 10^{-3} \text{ C}$. Potential $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} = \frac{9 \times 10^9 \times 1.45 \times 10^{-3}}{1.2} = 1.08 \times 10^7 \text{ V}$.
Solution: For $r < R$: $E = 0$ (graph is on x-axis), $V = \text{constant}$ (horizontal line). For $r > R$: $E \propto 1/r^2$ (steep curve), $V \propto 1/r$ (less steep curve). At $r=R$, $V$ and $E$ are maximum.
Solution: Potential at center of the ring $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$. Work done $W = qV = \frac{1}{4\pi\epsilon_0} \frac{qQ}{R}$.
Solution: $\vec{E} = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$. $\frac{\partial V}{\partial x} = -2xy - z^3$. $\frac{\partial V}{\partial y} = -x^2$. $\frac{\partial V}{\partial z} = -3xz^2$. Therefore, $\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$.
Solution: Yes. At any point on the equatorial plane of an electric dipole, the potential $V = 0$ because the distances to $+q$ and $-q$ are equal. However, the electric field $\vec{E}$ is not zero; it points anti-parallel to the dipole moment.
Solution: Distance from center to each corner $r = \frac{\sqrt{(\sqrt{2})^2 + (\sqrt{2})^2}}{2} = \frac{2}{2} = 1 \text{ m}$. $V = \frac{k}{r} \sum q = 9 \times 10^9 \times (100 - 50 + 20 - 60) \times 10^{-6} / 1 = 9 \times 10^9 \times 10 \times 10^{-6} = 9 \times 10^4 \text{ V}$.
Solution: The point ($0.1\text{m}$) lies inside the sphere. Potential inside is constant and equal to surface potential. $V = \frac{kQ}{R} = \frac{9 \times 10^9 \times 24 \times 10^{-6}}{0.2} = 1.08 \times 10^6 \text{ V}$.
Solution: $W = \oint \vec{E} \cdot d\vec{l} = 0$. This signifies that the electrostatic field is a conservative field, meaning work done is path-independent and only depends on initial and final positions.
Solution: Let point $P$ be at distance $r$ from center. Distances are $(r-a)$ from $+q$ and $(r+a)$ from $-q$. $V = V_+ + V_- = \frac{kq}{r-a} - \frac{kq}{r+a} = kq \left[ \frac{r+a - (r-a)}{r^2 - a^2} \right] = \frac{k(q \cdot 2a)}{r^2 - a^2} = \frac{kp}{r^2 - a^2}$. If $r \gg a$, $V \approx \frac{kp}{r^2}$.
Solution: For any point on the equatorial line at distance $r$ from center, the distance to $+q$ and $-q$ is $\sqrt{r^2 + a^2}$. Total potential $V = \frac{kq}{\sqrt{r^2+a^2}} + \frac{k(-q)}{\sqrt{r^2+a^2}} = 0$.
Solution: $p = q \times 2a = 8 \times 10^{-9} \times 0.04 = 3.2 \times 10^{-10} \text{ Cm}$. $\tau = pE\sin\theta \Rightarrow 4\sqrt{3} = pE\sin 60^\circ = pE(\sqrt{3}/2) \Rightarrow pE = 8$. $U = -pE\cos\theta = -8 \cos 60^\circ = -8(1/2) = -4 \text{ Joules}$.
Solution: $r_1 \approx r - a\cos\theta$, $r_2 \approx r + a\cos\theta$. $V = \frac{kq}{r_1} - \frac{kq}{r_2} = kq \frac{r_2 - r_1}{r_1 r_2} \approx kq \frac{2a\cos\theta}{r^2} = \frac{kp\cos\theta}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}$.
Solution: Distance to center is $a$. $V = 6 \times \frac{kQ}{a} = \frac{6Q}{4\pi\epsilon_0 a}$. Electric field vectors from opposite vertices cancel out perfectly in a regular hexagon. Thus, $E_{net} = 0$.
Solution: Volume conservation: $\frac{4}{3}\pi R_{new}^3 = 27 \times \frac{4}{3}\pi r^3 \Rightarrow R_{new} = 3r = 9 \text{ mm}$. Total charge $Q = 27q = 27 \times 10^{-12} \text{ C}$. $C = 4\pi\epsilon_0 R_{new} = \frac{9 \times 10^{-3}}{9 \times 10^9} = 1 \text{ pF}$. $V = \frac{Q}{C} = \frac{27 \times 10^{-12}}{10^{-12}} = 27 \text{ V}$.
Solution: Potential at inner shell is due to its own charge and the charge of the outer shell. $V_{inner} = \frac{k Q_1}{R} + \frac{k Q_2}{2R} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_1}{R} + \frac{Q_2}{2R} \right)$.
Solution: $r_A = \sqrt{1^2+2^2+3^2} = \sqrt{14}$. $r_B = \sqrt{1^2+1^2+(-1)^2} = \sqrt{3}$. $r_C = \sqrt{2^2+2^2+2^2} = \sqrt{12}$. Since $V \propto 1/r$, smaller distance means higher potential. $r_B < r_C < r_A$, therefore $V_B > V_C > V_A$.
Solution: The surfaces are close together between the charges. The plane exactly midway and perpendicular to the dipole axis (equatorial plane) is the locus of all points where $V=0$.
Solution: Let point be $(x,y)$ with $q$ at origin and $-2q$ at $(d,0)$. $V = \frac{kq}{\sqrt{x^2+y^2}} + \frac{k(-2q)}{\sqrt{(x-d)^2+y^2}} = 0 \Rightarrow \frac{1}{x^2+y^2} = \frac{4}{(x-d)^2+y^2}$. Solving this yields the equation of a circle. Locus is a spherical surface in 3D.
Solution: (1) No work is done moving a charge on it. (2) Electric field is perpendicular to it. They get closer near a charge because $E = -dV/dr$. Near the charge, $E$ is strong, so $dr$ (spacing for the same $dV$) must be small.
Solution: (i) Set of parallel planes perpendicular to the field. (ii) Two individual spheres around each charge that eventually merge into a single peanut-shaped surface enclosing both.
Solution: $dV = -\vec{E} \cdot d\vec{r} = -E dr \cos\theta$. For a given $dr$, the maximum decrease in $V$ (most negative $dV$) occurs when $\cos\theta = 1$, i.e., $\theta = 0$. This means displacement $d\vec{r}$ is parallel to $\vec{E}$.
Solution: The work done is exactly the same in both cases. Electrostatic force is conservative, so the work done depends only on the initial and final positions, not on the path taken.
Solution: $E_x = -\frac{dV}{dx} = -8x$. Faces of cube are at $x = \pm 0.5$. Flux $\Phi = E(0.5)A - E(-0.5)A = (-4)(1) - (4)(1) = -8$. By Gauss's Law, $Q_{enc} = \epsilon_0 \Phi = -8\epsilon_0$.
Solution: When identical spheres are connected, charge distributes equally to equalize potential. Total charge $= Q - 2Q = -Q$. Charge on each sphere $= -Q/2$. Final charge on A is $-Q/2$.
Solution: Work is path independent. Direct path A to B: $d\vec{r} = 2\hat{i}$. $W = \vec{F} \cdot d\vec{r} = q\vec{E} \cdot d\vec{r} = (2 \times 10^{-9})(20\hat{i}) \cdot (2\hat{i}) = 40 \times 10^{-9} \times 2 = 80 \times 10^{-9} \text{ J} = 8 \times 10^{-8} \text{ J}$.
Solution: Yes, $\nabla V$ is a vector. The relation $\vec{E} = -\nabla V$ has a negative sign which indicates that the electric field points in the direction of decreasing electric potential.
Solution: If a *volume* is an equipotential region (e.g., inside a conductor), then $dV=0$ implies $E=0$. However, on a single 2D equipotential *surface* with different potentials around it, $E \neq 0$.
Solution: Inside a hollow charged conductor, the potential is constant and equals the potential on its surface. Therefore, the potential at the center is $10\text{ V}$.
Solution: Work to bring $q_1$ is $W_1 = q_1 V(\vec{r}_1)$. Work to bring $q_2$ is against external field $W_2 = q_2 V(\vec{r}_2)$ PLUS work against $q_1$ which is $W_{12} = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}}$. Total $U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}}$.
Solution: $U_{initial} = \frac{k}{a} [(q)(2q) + (2q)(-3q) + (-3q)(q)] = \frac{k}{a} [2q^2 - 6q^2 - 3q^2] = -\frac{7kq^2}{a}$. Dissociation work $W = U_{final} - U_{initial} = 0 - (-7kq^2/a) = +\frac{7kq^2}{a}$.
Solution: Distance $r = 9 - (-9) = 18 \text{ cm} = 0.18 \text{ m}$. $U = \frac{k q_1 q_2}{r} = \frac{9 \times 10^9 \times 7 \times 10^{-6} \times (-2 \times 10^{-6})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \text{ Joules}$.
Solution: Work done $W = U_{final} - U_{initial} = U_{\infty} - U = 0 - (-0.7 \text{ J}) = +0.7 \text{ Joules}$.
Solution: Torque $\tau = pE\sin\theta$. Work done to rotate from $\theta_1$ to $\theta_2$ is $W = \int pE\sin\theta d\theta = pE(-\cos\theta_2 + \cos\theta_1)$. Assuming $U=0$ at $\theta_1 = 90^\circ$, potential energy at angle $\theta$ is $U(\theta) = -pE\cos\theta = -\vec{p} \cdot \vec{E}$.
Solution: Total $p = N_A \times 10^{-29} = 6.02 \times 10^{23} \times 10^{-29} = 6 \times 10^{-6} \text{ Cm}$. Initial $U_i = -pE\cos 0^\circ = -6\times 10^{-6} \times 10^6 = -6 \text{ J}$. Final $U_f = -pE\cos 60^\circ = -6(0.5) = -3 \text{ J}$. $\Delta U = -3 - (-6) = +3 \text{ J}$. Heat released is $3 \text{ J}$.
Solution: Four sides length $a$: two $-q(+q)$ pairs, one $-q,-q$ pair, one $+q,+q$ pair. Diagonals length $a\sqrt{2}$: two $-q(+q)$ pairs. $U = \frac{k}{a}[(-q^2) + (-q^2) + q^2 + q^2] + \frac{k}{a\sqrt{2}}[-q^2 - q^2] = 0 - \frac{2kq^2}{a\sqrt{2}} = -\sqrt{2}\frac{kq^2}{a}$.
Solution: Distance from corner to center is half the body diagonal: $r = \frac{L\sqrt{3}}{2}$. The potential at center due to 8 corners is $V = 8 \times \frac{kq}{r} = 8 \times \frac{kq}{L\sqrt{3}/2} = \frac{16kq}{L\sqrt{3}}$. $U = qV = \frac{16kq^2}{\sqrt{3}L}$.
Solution: (i) Stable: $\vec{p}$ is parallel to $\vec{E}$ ($\theta = 0^\circ$). $U = -pE$ (minimum energy). (ii) Unstable: $\vec{p}$ is anti-parallel to $\vec{E}$ ($\theta = 180^\circ$). $U = +pE$ (maximum energy).
Solution: $U = \frac{k (q)(q)}{r} = \frac{kq^2}{r}$. New $U' = \frac{k (q/2)(q/2)}{2r} = \frac{k (q^2/4)}{2r} = \frac{kq^2}{8r} = \frac{U}{8}$.
Solution: Principle: Capacitance of a conductor increases when an earthed conductor is brought near it. $C$ with slab: $V = E_0(d-t) + E_m t = \frac{\sigma}{\epsilon_0}(d-t) + \frac{\sigma}{K\epsilon_0}t = \frac{Q}{A\epsilon_0}[d-t+t/K]$. $C = \frac{Q}{V} = \frac{\epsilon_0 A}{d - t(1-1/K)}$.
Solution: $C_0 = \frac{\epsilon_0 A}{d} = 8\text{pF}$. New $C' = \frac{K\epsilon_0 A}{d/2} = 2K \left(\frac{\epsilon_0 A}{d}\right) = 2 \times 6 \times 8 = 96 \text{ pF}$.
Solution: Inside conductor $E=0$. $V = E_0(d-t) + 0 = \frac{\sigma}{\epsilon_0}(d-t) = \frac{Q}{A\epsilon_0}(d-t)$. $C = \frac{Q}{V} = \frac{\epsilon_0 A}{d-t} = \frac{\epsilon_0 A}{d(1-t/d)} = \frac{C_0}{1-t/d}$.
Solution: Alignment of dipole moments of a dielectric in an external electric field is polarization. It creates an induced internal field opposite to the external field, reducing the net field and potential difference ($V$). Since $C = Q/V$, a lower $V$ means higher capacitance ($C = KC_0$).
Solution: $U = \frac{1}{2}CV^2 = \frac{1}{2} \times (12 \times 10^{-12}) \times (50)^2 = 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8} \text{ Joules}$.
Solution: Potentials equalize: $V_a = V_b \Rightarrow \frac{k Q_a}{a} = \frac{k Q_b}{b} \Rightarrow \frac{Q_a}{Q_b} = \frac{a}{b}$. Ratio of densities $\frac{\sigma_a}{\sigma_b} = \frac{Q_a / (4\pi a^2)}{Q_b / (4\pi b^2)} = \frac{Q_a}{Q_b} \times \frac{b^2}{a^2} = \left(\frac{a}{b}\right) \left(\frac{b^2}{a^2}\right) = \frac{b}{a}$.
Solution: Charge $Q$ is constant. (i) $C$ increases to $KC$. (ii) $E$ decreases to $E/K$ (since $V=Q/C$ decreases, and $E=V/d$). (iii) $U = Q^2/2C$ decreases to $U/K$.
Solution: Potential $V$ is constant. $C$ increases to $KC$. (i) $Q = CV$ increases to $K \times Q$. (ii) $U = \frac{1}{2}CV^2$ increases to $K \times U$.
Solution: Inner shell $+Q$, outer $-Q$. Potential difference $V = V_a - V_b = \frac{Q}{4\pi\epsilon_0} (\frac{1}{a} - \frac{1}{b}) = \frac{Q}{4\pi\epsilon_0} (\frac{b-a}{ab})$. Capacitance $C = \frac{Q}{V} = \frac{4\pi\epsilon_0 ab}{b-a}$.
Solution: $C = 4\pi\epsilon_0 R = \frac{1}{9 \times 10^9} \times 6400 \times 10^3 = 7.11 \times 10^{-4} \text{ F} = 711 \mu\text{F}$.
Solution: Common potential $V = \frac{C_1V_1 + C_2V_2}{C_1+C_2}$. Initial $U_i = \frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2$. Final $U_f = \frac{1}{2}(C_1+C_2)V^2$. Loss $\Delta U = U_i - U_f = \frac{1}{2} \frac{C_1 C_2}{C_1+C_2} (V_1 - V_2)^2$. This energy is lost as heat and electromagnetic radiation in the connecting wires.
Solution: $C_1=600\text{pF}, V_1=200\text{V}, C_2=600\text{pF}, V_2=0$. $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1+C_2} (V_1 - 0)^2 = \frac{1}{2} (\frac{600 \times 600}{1200} \times 10^{-12}) (200)^2 = \frac{1}{2} (300 \times 10^{-12}) (40000) = 6 \times 10^{-6} \text{ J}$.
Solution: $C_p = C_1 + C_2 + C_3 = 2+3+4 = 9 \mu\text{F}$. Charges: $Q_1 = 2\mu\text{F} \times 100\text{V} = 200\mu\text{C}$; $Q_2 = 3 \times 100 = 300\mu\text{C}$; $Q_3 = 4 \times 100 = 400\mu\text{C}$.
Solution: Since all are equal ($10\mu\text{F}$), the bridge is balanced ($C_1/C_2 = C_3/C_4$). The central capacitor is ineffective. The circuit simplifies to two parallel branches, each having two $10\mu\text{F}$ in series. Series branch: $10/2 = 5\mu\text{F}$. Total $C_{eq} = 5\mu\text{F} + 5\mu\text{F} = 10\mu\text{F}$.
Solution: Initial total $U_i = 2 \times (\frac{1}{2}CV^2) = CV^2$. After switch open: Cap A (still connected) has $V'=V, C'=KC$, $U_A = \frac{1}{2}(KC)V^2$. Cap B (disconnected) has $Q'=CV, C'=KC$, $U_B = \frac{Q^2}{2(KC)} = \frac{(CV)^2}{2KC} = \frac{CV^2}{2K}$. Final $U_f = \frac{KCV^2}{2} + \frac{CV^2}{2K} = \frac{CV^2}{2}(K + \frac{1}{K})$. Ratio $U_i/U_f = 2 / (K + 1/K) = \frac{2K}{K^2+1}$.
Solution: Series branch $C_s = 10/3 \mu\text{F}$. Total $C_{eq} = 10/3 + 10 = 40/3 \mu\text{F}$. Charge on parallel cap $C_4$: $Q_4 = 10\mu\text{F} \times 500\text{V} = 5000\mu\text{C} = 5\text{mC}$. Charge on series branch (same for $C_1,C_2,C_3$): $Q_s = (10/3)\mu\text{F} \times 500\text{V} = 5000/3 \mu\text{C} \approx 1.67\text{mC}$.
Solution: Common potential $V = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$.
Solution: Loss $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1+C_2} V^2 = \frac{1}{2} \left( \frac{4\times 2}{4+2} \right) \times 10^{-6} \times (200)^2 = \frac{1}{2} (\frac{8}{6} \times 10^{-6}) (40000) = \frac{4}{6} \times 0.04 = 0.0267 \text{ J}$.
Solution: $u = \frac{1}{2}\epsilon_0 E^2$. For uniform field in capacitor volume $V_{ol} = Ad$, integral is $\int u \, dV_{ol} = u(Ad) = (\frac{1}{2}\epsilon_0 E^2)(Ad) = \frac{1}{2}(\frac{\epsilon_0 A}{d})(Ed)^2 = \frac{1}{2}CV^2 = U$.
Solution: $U_p = \frac{1}{2}C_p(100)^2 = 0.25 \Rightarrow C_1+C_2 = \frac{0.50}{10000} = 50\mu\text{F}$. $U_s = \frac{1}{2}C_s(100)^2 = 0.045 \Rightarrow \frac{C_1C_2}{C_1+C_2} = \frac{0.090}{10000} = 9\mu\text{F}$. So $C_1 C_2 = 9 \times 50 = 450$. $C_1, C_2$ are roots of $x^2 - 50x + 450 = 0$. Roots are $\frac{50 \pm \sqrt{2500 - 1800}}{2} = \frac{50 \pm \sqrt{700}}{2}$. Not perfect integers, approx $25 \pm 13.2$. Wait, usually these are perfect. Let's re-read: $0.25 = \frac{1}{2} (C_1+C_2) 10^4 \Rightarrow C_1+C_2 = 50 \mu\text{F}$. Series: $0.045 = \frac{1}{2} C_s 10^4 \Rightarrow C_s = 9\mu\text{F}$. $C_1C_2/50 = 9 \Rightarrow C_1C_2 = 450$. Let's assume standard values, e.g., $C_1, C_2$ could be $10$ and $45$ (sum $55 \neq 50$). Roots $25 \pm \sqrt{175} = 25 \pm 13.2 \mu\text{F} \approx 38.2 \mu\text{F}$ and $11.8 \mu\text{F}$.
