Electrostatics – JEE Main PhysicsPractice Questions & PYQs

${\text{V}}_{1}= {\text{V}}_{2}$ $\;{ {\text{kQ}}_{1}}/{8}=\;{ {\text{kQ}}_{2}}/{18} \;\; \{ {\text{v}}_{surface}=\;{ {\text{kQ}}}/{ {\text{R}}}\}$ $\;{ {\text{Q}}_{1}}/{ {\text{Q}}_{2}}=\;\frac{4}{9}$ $\;{ {\text{E}}_{ {\text{S}}_{1}}}/{ {\text{E}}_{ {\text{S}}_{2}}}=\;{ {\text{kQ}}_{1}}/{ {{\text{r}}^{2}}_{1}} \;{ {{\text{r}}^{2}}_{2}}/{ {\text{kQ}}_{2}} \{ {\text{E}}_{surface}=\;{ {\text{kQ}}}/{ {\text{R}}^{2}}\}$ $=\;{ {\text{Q}}_{1}}/{ {\text{Q}}_{2}} (\;{ {\text{r}}_{2}}/{ {\text{r}}_{1}}) ^{2}=\;\frac{4}{9} \times (\;\frac{9}{4}) ^{2}=\;\frac{9}{4}$

Let $r, q,$ and $V_s = \frac{kq}{r}$ be the radius, charge, and potential of one small bubble.
When three identical bubbles coalesce, volume conservation implies $\frac{4}{3}\pi R^3 = 3 \times (\frac{4}{3}\pi r^3)$, which yields $R = 3^{1/3}r$.
The total charge of the resultant bigger bubble is $Q = 3q$, so its potential is $V_b = \frac{kQ}{R} = \frac{k(3q)}{3^{1/3}r} = 3^{1 - 1/3} \left( \frac{kq}{r} \right) = 3^{2/3} V_s$.
The ratio of the potential of one initial bubble to that of the resultant bigger bubble is $\frac{V_s}{V_b} = \frac{V_s}{3^{2/3}V_s} = \frac{1}{3^{2/3}}$.
Thus, the required ratio is $1 : 3^{2/3}$.

Given $d=0.02 \, \text{m}$, $L=0.1 \, \text{m}$, $q=10^{-6} \, \text{C}$, $Q=24 \times 10^{-6} \, \text{C}$, and $k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$.
The linear charge density is $\lambda = \frac{Q}{L}$. The differential force on $q$ due to a charge element $dq = \lambda dx$ at distance $x$ is $dF = \frac{k q \lambda dx}{x^2}$.
Integrating over the wire from $x=d$ to $x=d+L$: $F = \int_{d}^{d+L} \frac{k q \lambda}{x^2} dx = k q \lambda \left[ -\frac{1}{x} \right]_{d}^{d+L} = k q \lambda \left( \frac{1}{d} - \frac{1}{d+L} \right)$.
Simplifying with $\lambda = \frac{Q}{L}$, we get $F = \frac{k q \lambda L}{d(d+L)} = \frac{k q Q}{d(d+L)}$.
Substituting the values: $F = \frac{(9 \times 10^9)(10^{-6})(24 \times 10^{-6})}{(0.02)(0.12)} = \frac{0.216}{0.0024} = 90 \, \text{N}$.

The resultant dipole moment $\vec{p}$ for a system of point charges is defined as $\vec{p} = \sum q_i \vec{r}_i$. Given the charges $q_1 = +2q$ at $\vec{r}_1 = -3a \hat{j}$, $q_2 = +3q$ at $\vec{r}_2 = 2a \hat{i}$, and $q_3 = -4q$ at $\vec{r}_3 = -2a \hat{i}$, we calculate the vector sum:
$\vec{p} = (2q)(-3a \hat{j}) + (3q)(2a \hat{i}) + (-4q)(-2a \hat{i})$
Expanding this expression gives:
$\vec{p} = -6qa \hat{j} + 6qa \hat{i} + 8qa \hat{i}$
Grouping the components yields:
$\vec{p} = (6 + 8)qa \hat{i} - 6qa \hat{j} = 14qa \hat{i} - 6qa \hat{j}$
Factoring out $2qa$, we obtain the resultant dipole moment:
$\vec{p} = 2qa(7 \hat{i} - 3 \hat{j})$

Assertion ⟶ True : (in electrostatics, conductor can have charge only on its surface)Reason ⟶ True : [inside capacitor, electric field exists therefore charges can improve force] But reason is not correct explanation of assertion

The relationship between the electric field $\vec{E}$ and potential $V$ is given by $dV = -\vec{E} \cdot d\vec{r} = -(Ax \, dx + By \, dy)$.
Integrating from the origin $(0,0)$ to a point $(x,y)$, we obtain $V(x, y) - V(0,0) = -\int_{0}^{x} Ax \, dx - \int_{0}^{y} By \, dy = -\frac{Ax^2}{2} - \frac{By^2}{2}$.
Rearranging for the potential at the origin, we have $V(0,0) = V(x, y) + \frac{Ax^2}{2} + \frac{By^2}{2}$.
Substituting the given values $V(10, 20) = 500 \, \text{V}$, $x=10$, $y=20$, $A=10 \, \text{V/m}^2$, and $B=5 \, \text{V/m}^2$:
$V(0,0) = 500 + \frac{10(10)^2}{2} + \frac{5(20)^2}{2} = 500 + 500 + 1000 = 2000 \, \text{V}$.

At equilibrium, the forces acting on the bob are gravity $mg$ (downward), the electric force $qE$ (horizontal), and the tension $T$ directed along the string.
Resolving the tension into perpendicular components, we balance the forces as $T \cos\theta = mg$ and $T \sin\theta = qE$, where $\theta$ is the angle the string makes with the vertical.
Squaring both equations gives $T^2 \cos^2\theta = m^2 g^2$ and $T^2 \sin^2\theta = q^2 E^2$.
Adding these equations yields $T^2 (\cos^2\theta + \sin^2\theta) = m^2 g^2 + q^2 E^2$.
Applying the identity $\sin^2\theta + \cos^2\theta = 1$, we get $T^2 = m^2 g^2 + q^2 E^2$.
Solving for $T$, we find the final tension to be $T = \sqrt{m^2 g^2 + q^2 E^2}$.

A is incorrect because electrostatic field lines originate from positive charges and terminate on negative ones, unlike magnetic field lines which form closed loops. B is correct because the electric field for a charge $q > 0$ is directed radially outward. C is correct because Gauss's Law, $\oint \vec{E} \cdot d\vec{A} = q/\epsilon_0$, relies fundamentally on the inverse-square nature of the Coulomb force. D is correct because static electric fields are conservative, meaning $\oint \vec{E} \cdot d\vec{l} = 0$, so the work done over a closed path is zero. E is correct because Coulomb's force is a central force ($\vec{F} \parallel \vec{r}$), which conserves angular momentum $\vec{L} = \vec{r} \times \vec{p}$, thereby restricting motion to a single plane.

$\;{\Delta {\text{V}}}/{(12 / \pi ) \times 10^{-2}} \times 5 \times 10^{-9}=\;\frac{4}{3} \pi ( {\text{r}}^{3}) \rho ( {\text{g}})$ $\Delta {\text{V}} \times 5 \times 10^{-7}=16 ( {\text{r}}^{3}) \rho ( {\text{g}})$ $\Delta {\text{V}}=480, {\text{V}}_{ {\text{A}}}< {\text{V}}_{ {\text{B}}}$Plate A will be lower potential

The Poisson equation relates the electrostatic potential $V$ to the volume charge density $\rho$ as $\rho = -\epsilon_0 \nabla^2 V$. For spherical symmetry, $\nabla^2 V = \frac{1}{r^2} \frac{d}{dr}(r^2 \frac{dV}{dr})$.
Given $V = a r^3 + b$, we have $\frac{dV}{dr} = 3ar^2$, so $\nabla^2 V = \frac{1}{r^2} \frac{d}{dr}(3ar^4) = 12ar$.
Thus, the charge density is $\rho = -12ar\epsilon_0$. The total charge $Q$ within radius $r=1$ is calculated as $Q = \int_{0}^{1} \rho (4\pi r^2) dr$.
Substituting $\rho$, $Q = \int_{0}^{1} (-12ar\epsilon_0)(4\pi r^2) dr = -48\pi a \epsilon_0 \int_{0}^{1} r^3 dr$.
Evaluating the integral, $Q = -48\pi a \epsilon_0 \left[\frac{r^4}{4}\right]_{0}^{1} = -12\pi a \epsilon_0$.
Comparing this with the given expression $Q = \alpha \pi a \epsilon_0$, we find $\alpha = -12$.

The potential at a distance $r$ from a conducting spherical shell of radius $R$ with charge $Q$ is $\frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$ for $r \ge R$ and $\frac{1}{4 \pi \epsilon_0} \frac{Q}{R}$ for $r < R$.
For shell A ($r=a$), the potential is the sum of contributions from all three shells: $V_A = \frac{1}{4 \pi \epsilon_0} (\frac{q_1}{a} + \frac{q_2}{b} + \frac{q_3}{c})$.
For shell B ($r=b$), since $a < b < c$, $q_1$ acts as if it is at its surface, $q_2$ is at its surface, and $q_3$ is outside: $V_B = \frac{1}{4 \pi \epsilon_0} (\frac{q_1}{b} + \frac{q_2}{b} + \frac{q_3}{c}) = \frac{1}{4 \pi \epsilon_0} (\frac{q_1+q_2}{b} + \frac{q_3}{c})$.
For shell C ($r=c$), all charges are effectively at or inside the surface $r=c$: $V_C = \frac{1}{4 \pi \epsilon_0} (\frac{q_1}{c} + \frac{q_2}{c} + \frac{q_3}{c}) = \frac{1}{4 \pi \epsilon_0} (\frac{q_1+q_2+q_3}{c})$.
These expressions match the formulas provided in option D.

${\text{V}}=5 ( {\text{x}}^{2}- {\text{y}}^{2}) {\text{V}}$ $\epsilon _{x}=\;\frac{-d v}{d x}=-10 x {i}^{∧}$ $\epsilon _{y}=\;\frac{-d v}{d x}=+10 y {i}^{∧} \Rightarrow \epsilon (2,3)=-20 {i}^{∧}+30 {j}^{∧}$

The distance of points $A$ and $B$ from the origin are $r_A = \sqrt{4^{2+}4^{2+}2^2} = 6 \, \text{m}$ and $r_B = \sqrt{2^{2+}2^{2+}1^2} = 3 \, \text{m}$.
The work done $W$ is given by $W = q(V_B - V_A) = kQq \left( \frac{1}{r_B} - \frac{1}{r_A} \right)$.
Substituting the given values, we have $W = (9 \times 10^9) (10^{-8}) (2 \times 10^{-6}) \left( \frac{1}{3} - \frac{1}{6} \right)$.
Simplifying the expression, $W = (18 \times 10^{-5}) \left( \frac{1}{6} \right) = 3 \times 10^{-5} \, \text{J}$.
Converting to the required units, $W = 30 \times 10^{-6} \, \text{J}$.

The electrostatic potential due to the external field $E = \frac{A}{r^2} \hat{r}$ is $V(r) = \int_{r}^{\infty } E \cdot dr = \frac{A}{r} = \frac{9 \times 10^5}{r} \, \text{V}$.
The potential energy due to the external field is $U_{ext} = q_1 V(r_1) + q_2 V(r_2)$:
$U_{ext} = (7 \times 10^{-6} \, \text{C})\left(\frac{9 \times 10^5}{0.09 \, \text{m}}\right) + (-2 \times 10^{-6} \, \text{C})\left(\frac{9 \times 10^5}{0.09 \, \text{m}}\right) = 70 \, \text{J} - 20 \, \text{J} = 50 \, \text{J}$
The interaction energy between the charges is $U_{\int} = \frac{k q_1 q_2}{d}$, where $d = 0.18 \, \text{m}$:
$U_{\int} = \frac{(9 \times 10^9)(7 \times 10^{-6})(-2 \times 10^{-6})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \, \text{J}$
The total electrostatic energy of the configuration is $U = U_{ext} + U_{\int} = 50 \, \text{J} - 0.7 \, \text{J} = 49.3 \, \text{J}$.

Distance from each dipole center to P: r=40 cm=0.4m Dipole moments: $p_A=(2\times10^{-6})(10^{-2})=2\times10^{-8}$ $p_B=(4\times10^{-6})(10^{-2})=4\times10^{-8}$ For dipole A (axial point), $E_A=\frac{1}{4\pi\epsilon_0}\frac{2p_A}{r^3}$ $=9\times10^9\cdot\frac{2(2\times10^{-8})}{(0.4)^3}$ =5625 For dipole B (equatorial point), $E_B=\frac{1}{4\pi\epsilon_0}\frac{p_B}{r^3}$ $=9\times10^9\cdot\frac{4\times10^{-8}}{(0.4)^3}$ =5625 So magnitudes are equal. Directions: $E_A$​ is horizontal $E_B$​ is vertical They are perpendicular. Hence resultant: $E=\sqrt{E_A^{2+}E_B^2}$ $=5625\sqrt{2}$ $=\frac{9}{16}\times10^4\sqrt{2}$

The charges are located at $30^{\circ}, 90^{\circ}, 150^{\circ}, 210^{\circ}, 270^{\circ}, \text{and } 330^{\circ}$ on the circle. Pairs at $(30^{\circ}, 210^{\circ})$ with charges $(+Q, +Q)$ and $(90^{\circ}, 270^{\circ})$ with charges $(+Q, +Q)$ create electric fields that cancel out due to symmetry and equal magnitudes. The remaining pair consists of a $-Q$ charge at $150^{\circ}$ and a $+Q$ charge at $330^{\circ}$. The field from the $+Q$ charge at $330^{\circ}$ points away from it (towards $150^{\circ}$), and the field from the $-Q$ charge at $150^{\circ}$ points towards it (towards $150^{\circ}$), so they add constructively. The resultant field is $\vec{E} = 2 \times \left( \frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} \right) \hat{r}_{150^{\circ}}$, where $\hat{r}_{150^{\circ}} = \cos(150^{\circ}) \hat{i} + \sin(150^{\circ}) \hat{j} = -\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}$. Thus, $\vec{E} = \frac{2Q}{4 \pi \epsilon_0 R^2} \left( -\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) = -\frac{Q}{4 \pi \epsilon_0 R^2} (\sqrt{3} \hat{i} - \hat{j})$.

The potential $V$ at the third corner of the equilateral triangle due to charges $q_1$ and $q_2$ is:
$V = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{r} \right) = \frac{k(q_1 + q_2)}{r}$
The work done $W$ to bring charge $q_3$ from infinity to that point is $W = q_3 \times V = \frac{k q_3 (q_1 + q_2)}{r}$.
Given $k = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$, $q_1 = 1 \, \text{nC}$, $q_2 = 2 \, \text{nC}$, $q_3 = 3 \, \text{nC}$, and $r = 3 \, \text{cm} = 0.03 \, \text{m}$:
$W = \frac{(9 \times 10^9) \times (3 \times 10^{-9}) \times (1 \times 10^{-9} + 2 \times 10^{-9})}{0.03}$
$W = \frac{9 \times 10^9 \times 3 \times 10^{-9} \times 3 \times 10^{-9}}{0.03} = \frac{81 \times 10^{-9}}{0.03} = 2700 \times 10^{-9} \, \text{J}$
$W = 2.7 \times 10^{-6} \, \text{J} = 2.7 \, \mu\text{J}$

${\text{E}}=\;{2 {\text{k}} \lambda }/{ {\text{R}}} \sin (\;\frac{\theta }{2})$ $100=2 \times \;\frac{9 \times 10^{9} \times Q}{\pi R^{2}}$ $Q =2.14 \times 10^{-9} {\text{C}}$

By Gauss's Law, the total electric flux $\Phi$ through the cube is $\frac{Q_{enclosed}}{\epsilon_0}$.
The charge $2q$ is placed at a vertex, which is shared by 8 such cubes; therefore, the contribution to the enclosed flux is $\Phi_1 = \frac{2q}{8\epsilon_0} = \frac{q}{4\epsilon_0}$.
The charge $q$ is placed at the center of a face, which is shared by 2 such cubes; therefore, the contribution to the enclosed flux is $\Phi_2 = \frac{q}{2\epsilon_0} = \frac{2q}{4\epsilon_0}$.
The total flux passing through the cube is the sum of these contributions: $\Phi = \Phi_1 + \Phi_2 = \frac{q}{4\epsilon_0} + \frac{2q}{4\epsilon_0} = \frac{3q}{4\epsilon_0}$.

$\tan 60^{\circ}=\;\frac{\;\frac{K P_2}{r^3}}{\frac{2 K P_1}{r_3}}=\;\frac{P_2}{2 P_1}={\sqrt{3}}$ $=\;\frac{P_2}{P_1}=2 {\sqrt{3}}$