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Ans: At equilibrium, $T \sin\theta = F_e = \frac{kq^2}{(2l\sin\theta)^2}$ and $T \cos\theta = mg$. Dividing gives $\tan\theta = \frac{kq^2}{4l^2\sin^2\theta \cdot mg}$. Thus, $q = 2l\sin\theta \sqrt{\frac{mg\tan\theta}{k}} = 2l\sin\theta \sqrt{4\pi\epsilon_0 mg\tan\theta}$.

Ans: Without gravity ($g=0$), the only forces are tension and electrostatic repulsion. The balls will repel to the maximum distance, $180^\circ$ apart. $2\theta = 180^\circ$. Tension $T = F_e = \frac{kq^2}{(2l)^2} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{4l^2}$.

Ans: Restoring force $F = -2F_e \cos\theta = -2\left(\frac{kQq}{x^2+a^2}\right)\frac{x}{\sqrt{x^2+a^2}}$. For $x \ll a$, $F \approx -\left(\frac{2kQq}{a^3}\right)x$. Since $F \propto -x$, it's SHM. $k_{eq} = \frac{2kQq}{a^3}$. Time period $T = 2\pi\sqrt{\frac{m}{k_{eq}}} = 2\pi\sqrt{\frac{ma^3}{2kQq}}$.

Ans: Time to travel distance $x$ is $t = x/v_0$. Acceleration in y-direction $a_y = \frac{qE}{m}$. Deflection $y = \frac{1}{2}a_y t^2 = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{x}{v_0}\right)^2 = \frac{qEx^2}{2mv_0^2}$.

Ans: Let $Q$ be at distance $x$ from $4q$. For $Q$ in eq: $\frac{k(4q)Q}{x^2} = \frac{kqQ}{(l-x)^2} \implies \frac{2}{x} = \frac{1}{l-x} \implies x = \frac{2l}{3}$. For $q$ in eq: $\frac{k(4q)(q)}{l^2} + \frac{kQq}{(l/3)^2} = 0 \implies \frac{4q}{l^2} + \frac{9Q}{l^2} = 0 \implies Q = -4q/9$.

Ans: The charge $Q$ is negative, placed between two positive charges. If displaced axially, the attractive force from the nearer positive charge increases, pulling it further away. Hence, equilibrium is Unstable axially (but stable transversely).

Ans: Effective acceleration due to gravity $g_{eff} = g + \frac{qE}{m}$ (since $qE$ acts downwards). Time period $T = 2\pi\sqrt{\frac{L}{g_{eff}}} = 2\pi\sqrt{\frac{L}{g + qE/m}}$.

Ans: $g_{eff} = \sqrt{g^2 + (qE/m)^2}$ (vector sum of perpendicular accelerations). $T = 2\pi\sqrt{\frac{L}{(g^2 + (qE/m)^2)^{1/2}}}$.

Ans: Apply Work-Energy Theorem. $W = \Delta K = - \Delta U$. $U_{initial} = \frac{kQq}{\sqrt{x_0^2 + d^2}}$, $U_{final} = \frac{kQq}{\sqrt{x^2 + d^2}}$. $\frac{1}{2}mv^2 = kQq \left[ \frac{1}{\sqrt{x_0^2 + d^2}} - \frac{1}{\sqrt{x^2 + d^2}} \right]$. Solve for $v$.

Ans: Initial K.E. = Final P.E. $\implies \frac{1}{2}m_e v_0^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r_{min}}$. Thus, $r_{min} = \frac{e^2}{2\pi\epsilon_0 m_e v_0^2}$.

Ans: Consider element $dq$. $dE = \frac{k dq}{r^2 + x^2}$. By symmetry, perpendicular components cancel. $dE_x = dE \cos\theta = \frac{k dq}{r^2 + x^2} \frac{x}{\sqrt{R^2 + x^2}}$. Integrating $dq$ yields $Q$. Thus, $E = \frac{kQx}{(R^2 + x^2)^{3/2}}$.

Ans: Set $dE/dx = 0$. Using quotient rule on $x(R^2+x^2)^{-3/2}$ yields $x = \pm R/\sqrt{2}$. Substituting this back into the E-field equation gives $E_{max} = \frac{2kQ}{3\sqrt{3}R^2}$.

Ans: $dE = \frac{k dq}{R^2} = \frac{k (\lambda R d\theta)}{R^2}$. Integrating the vertical components ($dE \sin\theta$) from $0$ to $\pi$: $E = \int_0^\pi \frac{k\lambda}{R} \sin\theta d\theta = \frac{k\lambda}{R} [-\cos\theta]_0^\pi = \frac{2k\lambda}{R}$.

Ans: $dE = \frac{k(\lambda dx)}{x^2}$. Integrate from $x = a$ to $x = a+L$. $E = k\lambda \int_a^{a+L} x^{-2} dx = k\lambda \left[ -\frac{1}{x} \right]_a^{a+L} = k\lambda \left( \frac{1}{a} - \frac{1}{a+L} \right) = \frac{kQL}{L(a)(a+L)} = \frac{kQ}{a(a+L)}$.

Ans: Using $dE_y = dE \cos\theta$. $E = \int_{-L}^{L} \frac{k \lambda dx}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}$. Let $x = y \tan\theta$, $dx = y \sec^2\theta d\theta$. Limits $-\theta_0$ to $\theta_0$ where $\sin\theta_0 = L/\sqrt{L^2+y^2}$. Result: $E = \frac{2k\lambda}{y} \sin\theta_0 = \frac{2k\lambda}{y} \frac{L}{\sqrt{L^2+y^2}}$.

Ans: As $L \to \infty$, the term $\frac{L}{\sqrt{L^2+y^2}} \to 1$. Thus, $E = \frac{2k\lambda}{y}$, which is the standard infinite wire formula ($\frac{\lambda}{2\pi\epsilon_0 y}$).

Ans: Integrate rings of radius $r$, thickness $dr$, charge $dq = \sigma 2\pi r dr$. $dE = \frac{k (dq) x}{(r^2+x^2)^{3/2}}$. $E = \int_0^R \frac{k (\sigma 2\pi r dr) x}{(r^2+x^2)^{3/2}} = \frac{\sigma x}{2\epsilon_0} \left[ \frac{-1}{\sqrt{r^2+x^2}} \right]_0^R = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{x}{\sqrt{R^2+x^2}} \right)$.

Ans: If $x \ll R$, the term $\frac{x}{\sqrt{R^2+x^2}} \approx \frac{x}{R} \approx 0$. Substituting into the disc formula gives $E = \frac{\sigma}{2\epsilon_0}$, which is identical to an infinite plane sheet.

Ans: Treat as elemental discs of thickness $dy$ and radius $r = \sqrt{R^2-y^2}$. Field of a disc at its center is $dE = \frac{dq}{2A\epsilon_0} = \frac{\rho dy}{2\epsilon_0}$. Integrating from $y=0$ to $R$: $E = \int_0^R \frac{\rho}{2\epsilon_0} dy = \frac{\rho R}{4\epsilon_0}$ (using exact spherical integration yields the same).

Ans: Due to the variable charge density, symmetric integration yields orthogonal components. Integrate $dE_x = - \frac{k \lambda_0 \cos\theta R d\theta}{R^2} \cos\theta$ and $dE_y = - \frac{k \lambda_0 \cos\theta R d\theta}{R^2} \sin\theta$ from $-\pi/2$ to $\pi/2$. Net field $E = \frac{k\lambda_0\pi}{2R}$ directed along negative x-axis.

Ans: Treat the square as one face of a cube of side $a$. The charge is exactly at the center of the cube. By Gauss's Law, total flux is $q/\epsilon_0$. Since the 6 faces are symmetric relative to the center, flux through one face is $\Phi = \frac{q}{6\epsilon_0}$.

Ans: The charge is shared by 2 cubes, so enclosed charge is $q/2$. Total flux for one cube is $q/2\epsilon_0$. The face containing the charge has zero flux (field lines parallel). The remaining 5 faces are not symmetric. Using solid angle math, flux through the opposite face is roughly $\sim 0.033 q/\epsilon_0$ (fraction depends on solid angle $\Omega \approx 0.41$).

Ans: An edge is shared by 4 identical cubes. The effective charge enclosed by one cube is $Q/4$. Thus, the total flux passing through the cube is $\Phi = \frac{Q}{4\epsilon_0}$.

Ans: Solid angle subtended by disc at distance $x$ is $\Omega = 2\pi(1 - \cos\theta)$ where $\cos\theta = \frac{x}{\sqrt{R^2+x^2}}$. Total flux in $4\pi$ steradians is $q/\epsilon_0$. Flux through disc $\Phi = \frac{\Omega}{4\pi} \frac{q}{\epsilon_0} = \frac{q}{2\epsilon_0} \left( 1 - \frac{x}{\sqrt{R^2+x^2}} \right)$.

Ans: For $r < R$: $q_{in} = \int_0^r \rho_0 \frac{r'}{R} 4\pi r'^2 dr' = \frac{4\pi\rho_0}{R} [\frac{r'^4}{4}]_0^r = \frac{\pi\rho_0 r^4}{R}$. Gauss Law: $E(4\pi r^2) = \frac{\pi\rho_0 r^4}{R\epsilon_0} \implies E = \frac{\rho_0 r^2}{4\epsilon_0 R}$. For $r > R$, total $Q = \pi\rho_0 R^3$, so $E = \frac{Q}{4\pi\epsilon_0 r^2} = \frac{\rho_0 R^3}{4\epsilon_0 r^2}$.

Ans: Gaussian cylinder of radius $r$, length $l$. $q_{in} = \rho (\pi r^2 l)$. Flux $\Phi = E(2\pi r l)$. By Gauss's Law: $E(2\pi r l) = \frac{\rho \pi r^2 l}{\epsilon_0} \implies E = \frac{\rho r}{2\epsilon_0}$.

Ans: E-field of sheet with hole is $E_{sheet} - E_{disc} = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0}(1 - \frac{x}{\sqrt{R^2+x^2}}) = \frac{\sigma x}{2\epsilon_0\sqrt{R^2+x^2}}$. For $x \ll R$, $E \approx \frac{\sigma x}{2\epsilon_0 R}$. Restoring force $F = -qE \approx -(\frac{q\sigma}{2\epsilon_0 R})x$. SHM frequency $f = \frac{1}{2\pi} \sqrt{\frac{q\sigma}{2\epsilon_0 m R}}$.

Ans: To maintain $E=0$ inside the conductor, a charge $-q$ is induced on inner surface $R_1$. To maintain overall neutrality, $+q$ is induced on outer surface $R_2$. $\sigma_{inner} = \frac{-q}{4\pi R_1^2}$, $\sigma_{outer} = \frac{+q}{4\pi R_2^2}$.

Ans: For $r < R_1$, $E = \frac{kq}{r^2}$ (inverse square drop). For $R_1 \le r \le R_2$, $E = 0$ (inside conductor). For $r > R_2$, $E = \frac{kq}{r^2}$ (resumes inverse square drop from a lower starting value).

Ans: Use superposition: Solid sphere of $+\rho$ and cavity sphere of $-\rho$. Field vector at point $\vec{r}$ inside cavity is $\vec{E} = \vec{E}_+ + \vec{E}_- = \frac{\rho \vec{r}}{3\epsilon_0} - \frac{\rho (\vec{r} - \vec{d})}{3\epsilon_0} = \frac{\rho \vec{d}}{3\epsilon_0}$. Since $\vec{d}$ is a constant vector, the field is perfectly uniform.

Ans: Field of $p_1$ at distance $r$ is $E_1 = \frac{2kp_1}{r^3}$. Force on $p_2$ is $F = p_2 \frac{dE_1}{dr} = p_2 \frac{d}{dr}(\frac{2kp_1}{r^3}) = - \frac{6k p_1 p_2}{r^4}$. Magnitude is $\frac{6k p_1 p_2}{r^4}$ (attractive if parallel).

Ans: Based on the derivation in Q31, $F \propto 1/r^4$. Thus, $n = 4$. Dipole-dipole force falls off much faster than point charge force ($1/r^2$).

Ans: Field of $p_1$ at equatorial distance $r$ is $\vec{E}_1 = -\frac{k\vec{p}_1}{r^3}$ (antiparallel to $\vec{p}_1$). Torque $\vec{\tau} = \vec{p}_2 \times \vec{E}_1$. Since $\vec{p}_2$ is parallel to $\vec{p}_1$, the angle between $\vec{p}_2$ and $\vec{E}_1$ is $180^\circ$. $\tau = p_2 E_1 \sin(180^\circ) = 0$.

Ans: The field has radial component $E_r = \frac{2kp\cos\theta}{r^3}$ and tangential component $E_\theta = \frac{kp\sin\theta}{r^3}$. The angle $\alpha$ made by $\vec{E}$ with the radial vector is $\tan\alpha = \frac{E_\theta}{E_r} = \frac{1}{2}\tan\theta$. Angle with x-axis is $\theta + \alpha = \theta + \tan^{-1}(\frac{\tan\theta}{2})$.

Ans: $F = \vec{p} \cdot \nabla \vec{E} = p \frac{dE}{dx}$ (since 1D). $E(x) = \alpha x \implies dE/dx = \alpha$. Net Force $F = p \alpha \hat{i}$.

Ans: Restoring torque $\tau = -pE \sin\theta \approx -pE\theta$. Equation of motion $I\alpha = -pE\theta \implies \alpha = -(\frac{pE}{I})\theta$. This is SHM with $\omega^2 = \frac{pE}{I}$. Time period $T = 2\pi\sqrt{\frac{I}{pE}}$.

Ans: At any instant, it forms a dipole of moment $p = q(2R)$ pointing from $-q$ to $+q$. As they rotate, the vector $\vec{p}$ rotates continuously. Since the integral of $\cos(\omega t)$ and $\sin(\omega t)$ over a full cycle is zero, the time-averaged dipole moment is zero.

Ans: $W = \Delta U = U_{final} - U_{initial} = (-pE\cos 180^\circ) - (-pE\cos 60^\circ) = pE - (-0.5 pE) = 1.5 pE = \frac{3}{2} pE$.

Ans: The electric field at the exact center of a uniformly charged ring is strictly zero ($E=0$). Therefore, the torque $\tau = pE\sin\theta = p(0)\sin\theta = 0$, regardless of its orientation.

Ans: A short dipole consists of equal and opposite charges very close to each other. Both charges are situated at $R/2$ (well inside the sphere of radius $R$). Net enclosed charge $Q_{in} = +q - q = 0$. Hence, net flux $\Phi = 0$.

Ans: Null point lies between them. $\frac{k(2)}{x^2} = \frac{k(8)}{(15-x)^2} \implies \frac{1}{x} = \frac{2}{15-x} \implies 15-x = 2x \implies 3x = 15 \implies x = 5\text{ cm}$. 5

Ans: The point $5\text{ cm}$ is inside the shell ($r < R$). The electric field inside a uniformly charged spherical shell is zero. Thus $Y = 0$. 0

Ans: At $r=1.5a$ (between spheres), enclosed charge is $Q$. $E_1 = \frac{kQ}{(1.5a)^2} = \frac{4kQ}{9a^2}$. At $r=4a$ (outside), enclosed charge is $Q - 2Q = -Q$. $E_2 = \frac{k|-Q|}{(4a)^2} = \frac{kQ}{16a^2}$. Ratio $E_1/E_2 = \frac{4/9}{1/16} = 64/9 \approx 7.11$. 64/9

Ans: Centripetal force $mv^2/r = eE = e(\frac{\lambda}{2\pi\epsilon_0 r}) = e(\frac{2k\lambda}{r})$. K.E. = $\frac{1}{2}mv^2 = \frac{1}{2}(e \cdot 2k\lambda) = ek\lambda = (1.6 \times 10^{-19}) \times (9 \times 10^9) \times (4 \times 10^{-6}) = 57.6 \times 10^{-16} = 5.76 \times 10^{-15}\text{ J}$. 5.76

Ans: Total flux $= q/\epsilon_0 = (8.85 \times 10^{-6}) / (8.85 \times 10^{-12}) = 10^6$. Flux per face $= 10^6 / 6 \approx 1.666 \times 10^5$. $Z = 10/6 = 5/3 \approx 1.67$. 1.67

Ans: $F \propto |5 \times -1| = 5$. After touching, $q_{new} = (5-1)/2 = 2\text{ mC}$. $F' \propto |2 \times 2| = 4$. Ratio $N = F/F' = 5/4 = 1.25$. 1.25

Ans: $qE = mg \implies E = \frac{mg}{q} = \frac{10^{-6} \times 10}{10^{-6}} = 10\text{ V/m}$. 10

Ans: $E_{axial} = \frac{2kp}{r^3} = \frac{2 \times (9 \times 10^9) \times (4 \times 10^{-9})}{(0.2)^3} = \frac{72}{0.008} = 9000 = 9 \times 10^3$. $E = 9$. 9

Ans: Area vector $\vec{A} = 1 \hat{i}$. $\vec{E}$ at $x=2$ is $20\hat{i} + 20y\hat{j}$. $\Phi = \int \vec{E} \cdot d\vec{A} = \int (20\hat{i} + 20y\hat{j}) \cdot (dA \hat{i}) = \int 20 dA = 20 \times 1 = 20$. 20

Ans: $K_{new} = 2K$. $F_{new} = \frac{F}{2}$. Decrease $= F - F/2 = F/2$, which is $50\%$ of the original force. $P = 50$. 50

Ans: Between plates: $E_{in} = \sigma/\epsilon_0$. Outside plates: $E_{out} = 0$. Ratio $X = E_{in} / 0 = \infty$. Infinity

Ans: By symmetry, the electric fields produced by opposite sides of the square cancel each other out perfectly at the geometric center. $E = 0$. 0

Ans: $r' = 1.25 r = \frac{5}{4}r$. $F' = \frac{F}{(5/4)^2} = F \times \frac{16}{25} = 100 \times \frac{16}{25} = 64\text{ N}$. 64

Ans: At equilibrium, $\tan\theta = \frac{qE}{mg}$. $\tan 45^\circ = 1$. Thus $qE/mg = 1$. 1

Ans: Inside a conductor, net $E = 0$. $E_{net} = E_{q} + E_{induced} = 0$. $E_{induced} = -E_{q}$. At center, $E_q = \frac{kq}{R^2}$. So magnitude of induced field is $\frac{kq}{R^2}$. Hence $x = 2$. 2

Ans: Null point must be outside the charges, closer to smaller charge ($-q$). Let it be at distance $d$ to the right of $a$. $\frac{k(4q)}{(a+d)^2} = \frac{k(q)}{d^2} \implies \frac{2}{a+d} = \frac{1}{d} \implies 2d = a+d \implies d=a$. Coordinate $x = a+a = 2a$. 2a

Ans: $E(r) = \frac{1}{4\pi\epsilon_0 r^2} \int_0^r \rho_0(1 - r'/R) 4\pi r'^2 dr' = \frac{\rho_0}{\epsilon_0 r^2} (\frac{r^3}{3} - \frac{r^4}{4R}) = \frac{\rho_0}{\epsilon_0} (\frac{r}{3} - \frac{r^2}{4R})$. To find max, $dE/dr = 0 \implies \frac{1}{3} - \frac{2r}{4R} = 0 \implies \frac{r}{2R} = \frac{1}{3} \implies r = \frac{2R}{3}$. 2/3 R

Ans: $\tau_{max} = pE = 10$. Work $W = \Delta U = U(180) - U(0) = pE - (-pE) = 2pE = 2(10) = 20\text{ Joules}$. 20

Ans: $q_{in} = \sigma 4\pi r^2, q_{out} = \sigma 4\pi R^2$. $q_{in} + q_{out} = Q \implies \sigma 4\pi (r^2+R^2) = Q \implies \sigma 4\pi = \frac{Q}{r^2+R^2}$. $q_{in} = (\frac{Q}{r^2+R^2}) r^2$. Qr² / (r² + R²)

Ans: The length of the wire inside the cube is equal to the side length $L$. Charge enclosed $q_{in} = \lambda L$. By Gauss's Law, Flux $\Phi = \frac{\lambda L}{\epsilon_0}$. λL / ε₀