Vardaan Learning Institute

Ans: At the macroscopic level, charges involved are usually in micro-coulombs ($\mu\text{C}$) or higher. Since $1\text{ C} \approx 6.25 \times 10^{18}$ electrons, adding or removing a few electrons makes an infinitesimally small fractional change. Thus, charge behaves like a continuous quantity.

Ans: $n = \frac{q}{e} = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.875 \times 10^{12}$ electrons. Since polythene is negatively charged, electrons were transferred from wool to polythene.

Ans: Yes, electrons have mass. Mass transferred = $n \times m_e = (1.875 \times 10^{12}) \times (9.1 \times 10^{-31}) \approx 1.7 \times 10^{-18}\text{ kg}$. (Negligibly small, but non-zero).

Ans: When touched, net charge is conserved and distributed equally (since identical). $Q_{net} = +Q + (-3Q) = -2Q$. Charge on each sphere = $Q_{net}/2 = -2Q / 2 = -Q$.

Ans: $-1.6 \times 10^{-12}\text{ C}$. Justification: Law of Conservation of Charge. The net charge of the isolated system (rod + silk) must remain zero.

Ans: Bring a negatively charged rod near the neutral sphere. Free electrons are repelled to the far side, leaving the near side positive. Ground the far side to let electrons escape to Earth. Remove the ground connection, then remove the rod. The sphere is left positively charged.

Ans: $n = \frac{q}{e} = \frac{0.32 \times 10^{-18}}{1.6 \times 10^{-19}} = 2$. Since $n$ is an integer ($2$), yes, this charge is physically possible.

Ans: For a continuous distribution, the algebraic sum becomes an integral over the length, area, or volume. $Q_{total} = \int dq = \int \lambda dl$ (or $\sigma dA$, or $\rho dV$).

Ans: $F = mg = (10 \times 10^{-6}\text{ kg}) \times 10\text{ m/s}^2 = 10^{-4}\text{ N}$.
By Coulomb's law: $F = k \frac{q^2}{r^2} \implies 10^{-4} = 9 \times 10^9 \frac{q^2}{(0.6)^2}$.
$q^2 = \frac{10^{-4} \times 0.36}{9 \times 10^9} = 4 \times 10^{-15} = 40 \times 10^{-16}$.
$q = \sqrt{40} \times 10^{-8} \approx 6.32 \times 10^{-8}\text{ C}$.

Ans: Position vector $\vec{r} = 3\hat{i} + 4\hat{j}$. Magnitude $r = \sqrt{3^2 + 4^2} = 5\text{ m}$. Unit vector $\hat{r} = \frac{3\hat{i} + 4\hat{j}}{5}$.
Magnitude $F = (9 \times 10^9) \frac{2 \times 10^{-6} \times 2 \times 10^{-6}}{5^2} = \frac{36 \times 10^{-3}}{25} = 1.44 \times 10^{-3}\text{ N}$.
Since $q_1, q_2$ have opposite signs, force is attractive (towards origin).
$\vec{F}_{12} = -F \hat{r} = -1.44 \times 10^{-3} (\frac{3\hat{i} + 4\hat{j}}{5})\text{ N}$.

Ans: The force in a medium is $F_m = \frac{F_{vacuum}}{K}$. Therefore, the new force is $F/80$.

Ans: Let separation be $2x$. $q$ is at $x$. For $q$ to be in equilibrium, net force on it is zero (always true if it's midway between identical charges). For $Q$ to be in equilibrium, force due to other $Q$ + force due to $q$ = $0$.
$\frac{k(Q)(Q)}{(2x)^2} + \frac{k(Q)(q)}{x^2} = 0 \implies \frac{Q^2}{4x^2} = -\frac{Qq}{x^2} \implies \frac{Q}{4} = -q \implies q = -Q/4$.

Ans: The graph is a straight line passing through the origin ($y = mx$, where $y=F, x=1/r^2$). The slope $m = \frac{1}{4\pi\epsilon_0} (q_1 q_2)$.

Ans: Let charges be at A, B, C. Force on C due to A and B: $F_A = F_B = \frac{kq^2}{a^2}$. Angle between them is $60^\circ$. Resultant $F_{net} = \sqrt{F_A^2 + F_B^2 + 2F_A F_B \cos 60^\circ} = \sqrt{F^2 + F^2 + 2F^2(1/2)} = \sqrt{3F^2} = \sqrt{3}\frac{kq^2}{a^2}$.

Ans: The electrostatic force drops to zero. For a metallic conductor, the dielectric constant $K = \infty$. $F_{medium} = F/K = F/\infty = 0$. (The induced charges on the metal completely cancel the external field inside).

Ans: $F_e/F_g = (kq^2/r^2) / (Gm^2/r^2) = kq^2/Gm^2$. Substituting values gives $F_e/F_g \approx 10^{36}$. The electrostatic force is immensely stronger and completely dominates at atomic scales.

Ans: Diagonals intersect at center $O$. Distance $r = a/\sqrt{2}$. Fields due to opposite corners ($+q$ and $+q$, or $-q$ and $-q$) cancel each other out due to symmetry. Therefore, the net electric field at the center is strictly Zero ($0$).

Ans: Acceleration $a = \frac{eE}{m_e} = \frac{(1.6 \times 10^{-19})(2.0 \times 10^4)}{9.1 \times 10^{-31}} \approx 3.5 \times 10^{15}\text{ m/s}^2$.
Using $s = \frac{1}{2}at^2 \implies t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2 \times 0.015}{3.5 \times 10^{15}}} \approx 2.9 \times 10^{-9}\text{ s}$ ($2.9\text{ ns}$).

Ans: To remain stationary, upward electric force must balance downward weight: $qE = mg$. Since $E$ is downward, $q$ must be negative to experience an upward force.
$|q| = \frac{mg}{E} = \frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4} = 3.3 \times 10^{-18}\text{ C}$.
Excess electrons $n = q/e = (3.3 \times 10^{-18}) / (1.6 \times 10^{-19}) \approx 20.6$ (Rounding to 21 electrons). The drop has an *excess* of electrons.

Ans: [Student should draw outward radial lines from both charges that curve away from each other along the perpendicular bisector]. The null point is exactly midway between the two identical charges, where $E_{net} = 0$.

Ans: If $E$ were not zero, free electrons inside the conductor would experience a force and move, creating a current. Since it's electrostatics (charges at rest), $E$ must be zero inside. Excess charges reside purely on the outer surface.

Ans: A null point is where net electric field is zero. For two opposite, unequal charges, the null point cannot lie between them. It lies on the line joining them, outside the region between them, closer to the charge with the smaller magnitude.

Ans: For fields to cancel exactly at the midpoint, the electric fields from both charges must point in opposite directions and have equal magnitude. Therefore, $q_1$ and $q_2$ must be identical in both magnitude and sign ($q_1 = q_2$).

Ans: Electric field lines represent the continuous force experienced by a test charge. A sudden break would imply that the electric force abruptly vanishes at that point and reappears, which physically does not happen in a charge-free region.

Ans: Let point $P$ be at distance $r$ from center. Dist to $\pm q$ is $\sqrt{r^2 + a^2}$. Magnitudes: $E_{+q} = E_{-q} = \frac{kq}{r^2+a^2}$. Vertical components ($E \sin\theta$) cancel. Horizontal components add up: $E_{eq} = 2 E_{+q} \cos\theta$.
$\cos\theta = \frac{a}{\sqrt{r^2+a^2}}$.
$E_{eq} = 2 \frac{kq}{r^2+a^2} \frac{a}{(r^2+a^2)^{1/2}} = \frac{k(q \cdot 2a)}{(r^2+a^2)^{3/2}} = \frac{kp}{(r^2+a^2)^{3/2}}$.
For short dipole ($r \gg a$): $E_{eq} \approx \frac{kp}{r^3}$.

Ans: $\tau = pE \sin\theta = (q \cdot 2a)E \sin\theta$.
$8\sqrt{3} = q \times (0.02) \times 10^5 \times \sin 60^\circ = q \times 2000 \times \frac{\sqrt{3}}{2} = 1000\sqrt{3} q$.
$q = \frac{8\sqrt{3}}{1000\sqrt{3}} = 8 \times 10^{-3}\text{ C} = 8\text{ mC}$.

Ans: $U = -pE \cos\theta$. We know $pE = \frac{\tau}{\sin\theta} = \frac{8\sqrt{3}}{\sqrt{3}/2} = 16$.
$U = -16 \times \cos 60^\circ = -16 \times (1/2) = -8\text{ Joules}$.

Ans: $\Phi = \vec{E} \cdot \vec{A} = (5\hat{i} + 4\hat{j} - \hat{k}) \cdot (2\hat{i} + 3\hat{j} + 0\hat{k})$.
$\Phi = (5 \times 2) + (4 \times 3) + (-1 \times 0) = 10 + 12 = 22\text{ V m}$ (or $\text{N m}^2/\text{C}$).

Ans: $E_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$.
$E_{eq} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$.
Therefore, $\frac{E_{axial}}{E_{eq}} = \frac{2p/r^3}{p/r^3} = 2 \implies E_{axial} = 2 E_{eq}$.

Ans: In a non-uniform field, the forces on $+q$ and $-q$ are unequal ($qE_1 \neq qE_2$). Thus, the net translational force is NOT zero. It will also generally experience a net torque unless $\vec{p}$ is perfectly parallel/antiparallel to $\vec{E}$.

Ans: By Gauss's law, since the uniform field lines entering the curved surface equal the field lines leaving the curved surface (no enclosed charge), the net flux over the closed hemispherical surface is Zero.

Ans: A charge at the corner is shared equally by 8 adjacent cubes meeting at that corner. The effective enclosed charge for one cube is $Q/8$. Therefore, flux through the cube = $\frac{Q}{8\epsilon_0}$.

Ans: Consider a spherical Gaussian surface of radius $r > R$ concentric with the shell. By symmetry, $E$ is radial and constant over this surface.
$\oint \vec{E} \cdot d\vec{A} = E \oint dA = E(4\pi r^2)$.
By Gauss's Law: $E(4\pi r^2) = \frac{Q_{enclosed}}{\epsilon_0}$. Here $Q_{enclosed} = Q$ (total charge on shell).
$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$.

Ans: Flux through inner shell ($S_1$) = $Q_1 / \epsilon_0$.
Flux through outer shell ($S_2$) encloses both charges = $(Q_1 + Q_2) / \epsilon_0$.
Ratio $\Phi_1 : \Phi_2 = \frac{Q_1}{Q_1 + Q_2}$.

Ans: No. Net flux zero only implies that the *net* enclosed charge is zero ($\sum q = 0$). For example, a closed surface enclosing an electric dipole has zero net flux, but the electric field $E$ at various points on the surface is definitely not zero.

Ans: The electric field from a large plane is perpendicular to the plane. If the circular area is *parallel* to the plane, its area vector is perpendicular to the plane, hence $\vec{E}$ and $\vec{A}$ are parallel ($\theta=0^\circ$).
$E = \sigma / 2\epsilon_0$. Area $A = \pi r^2 = \pi(0.5)^2 = 0.25\pi$.
$\Phi = EA = \frac{2.0 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}} \times 0.25\pi \approx 8.87 \times 10^4\text{ N m}^2/\text{C}$.

Ans: Volume charge density $\rho$ is the charge per unit volume. For a uniform solid sphere, $Q = \rho \times V = \rho \times (\frac{4}{3}\pi R^3)$.

Ans: Total length (circumference) of the ring $L = 2\pi a$. Linear charge density $\lambda = \frac{q}{L} = \frac{q}{2\pi a}$.

Ans: By Gauss's Law, the shape/size of the closed surface doesn't matter.
$\Phi = \frac{q}{\epsilon_0} = \frac{2.0 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 2.26 \times 10^5\text{ N m}^2/\text{C}$.

Ans: The total electric flux remains absolutely unchanged. Gauss's Law ($\Phi = q_{in}/\epsilon_0$) depends only on the total amount of charge enclosed, not on its exact location within the surface.

Ans: Choose a cylindrical Gaussian pillbox of cross-sectional area $A$ piercing the sheet. Flux only passes through the two flat end caps (since $E$ is parallel to curved surface).
$\Phi_{total} = EA + EA = 2EA$.
Charge enclosed $q_{in} = \sigma A$.
By Gauss's Law: $2EA = \frac{\sigma A}{\epsilon_0} \implies E = \frac{\sigma}{2\epsilon_0}$.

Ans: Between plates, fields from positive and negative plates point in the same direction.
$E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} \approx 1.9 \times 10^{-10}\text{ N/C}$.

Ans: In the outer regions, the electric fields produced by the two oppositely charged plates are in opposite directions but have the same magnitude ($\sigma/2\epsilon_0$). They cancel each other perfectly. Thus, $E_{outer} = 0$.

Ans: $E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{r} = (9 \times 10^9) \times \frac{2 \times (2 \times 10^{-4})}{0.10} = \frac{36 \times 10^5}{0.10} = 3.6 \times 10^7\text{ N/C}$.

Ans: Since it's a *conducting* sphere, all charge moves to the surface. It behaves exactly like a hollow shell. Graph shows $E=0$ from $r=0$ to $r=R$. At $r=R$, $E$ peaks. For $r>R$, $E \propto 1/r^2$.

Ans: No. The electric field of a line charge varies with distance ($E \propto 1/r$). As the particle moves, the field strength changes, hence the force ($qE$) changes, leading to a variable, non-constant acceleration.

Ans: Construct a spherical Gaussian surface inside the shell ($r < R$). Since all charge resides on the shell's outer surface, the charge enclosed by this Gaussian surface $q_{in} = 0$. By Gauss's Law, $\oint E \cdot dA = 0 \implies E = 0$.

Ans: Because the sheet is "infinite", the electric field lines generated are perfectly parallel and uniformly spaced. Since the lines never spread apart (diverge) or come together (converge), the flux density (field strength) remains constant regardless of distance.