Electrostatics Practice Questions

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 $1 : 3^{2/3}$.

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 $B=5 \, \text{V/m}^2$:
$V(0,0) = 500 + \frac{10(10)^2}{2} + \frac{5(20)^2}{2} = 500 + 500 + 1000 = 2000 \, \text{V}$.

$\;{\Delta {\text{V}}}/{(12 / \pi ) \times 10^{-2}} \times 5 \times 10^{-9}=\;\frac{4}{3} \pi ( {\text{r}}^{3}) \rho ( {\text{g}})$ $\Delta {\text{V}} \times 5 \times 10^{-7}=16 ( {\text{r}}^{3}) \rho ( {\text{g}})$ $\Delta {\text{V}}=480, {\text{V}}_{ {\text{A}}}< {\text{V}}_{ {\text{B}}}$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, $\vec{E} = \frac{2Q}{4 \pi \epsilon_0 R^2} \left( -\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) = -\frac{Q}{4 \pi \epsilon_0 R^2} (\sqrt{3} \hat{i} - \hat{j})$.

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

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