Electrostatics Practice Questions

Given, mass of each spheres, $m=10 {mg}=10 \times 10^{-3} {g}$ Length of thread $(l)=0.5 {m}$ Separation between charges, $d=0.2 {m}$ Charge of each sphere, $q=\;\frac{a}{21} \times 10^{-8} C$ Acceleration due to gravity $(g)=10 {ms}^{-2}$ The situation can be shown as below,Taking component of tension (T) $T \cos \theta \;\; =m g$. . . (i) $T \sin \theta \;\; =q E=\;\frac{k \mid q^{2}}{d^{2}}$ . . . (ii) $\sin \theta =\;\frac{d / 2}{l}=\;\frac{0.1}{0.5}=\;\frac{1}{5}$ $\cos \theta =\sqrt{1-\sin ^{2} \theta }=\sqrt{1-1 / 25}=\;\frac{\sqrt{24}}{5}$ $\therefore \;\; \tan \theta =\;\frac{\sin \theta }{\cos \theta }=\;\frac{k q^{2}}{d^{2} m g} \;\;$ [using Eqs. (i) and (ii)] $\Rightarrow \;\; \;\frac{1 / 5}{\sqrt{24} / 5}=\;\frac{k q^{2}}{d^{2} m g}$ $\Rightarrow \;\; \;\frac{1}{\sqrt{24}}=\;\frac{9 \times 10^{9} \times q^{2}}{(0.2)^{2} \times 10 \times 10^{-3} \times 10}$ $\Rightarrow \;\; q=\sqrt{\;\frac{(0.2)^{2} \times 10^{-1}}{\sqrt{24} \times 9 \times 10^{9}}} \Rightarrow q=3 \times 10^{-7} {C}$ $\Rightarrow \;\; \;\frac{a}{21} \times 10^{-8}=30 \times 10^{-8}$ $a=630$

The arrangement of charges is shown belowAs we know that, Coulomb's force between two charges. i.e. $q_{1}$ and $q_{2}$, $F=\;\frac{1}{4 \pi \epsilon _{0}} \;\frac{q_{1} q_{2}}{r^{2}}=\;\frac{1}{4 \pi \epsilon _{0}} \;\frac{q_{1} q_{2}}{(d^{2}+x^{2})}$. . . (i) Here, $q_{1}=q_{2}=q$ Force in ${SHM}, F=m \omega ^{2} x$ ...(ii) Since, in order to have ${SHM}+q$ should move downwards and force responsible for this will be only $F^{'}=F \sin \theta +F \sin \theta =2 F \sin \theta$. . . (iii) Using Eqs. (ii) and (iii), we get $2 F \sin \theta =m \omega ^{2} x$ $\Rightarrow \;\; \;\frac{2}{4 \pi \epsilon _{0}} \;\frac{q^{2}}{(d^{2}+x^{2})} \sin \theta =m \omega ^{2} x$ $\Rightarrow \;\; \;\frac{2}{4 \pi \epsilon _{0}} \;\frac{q^{2}}{(d^{2}+x^{2})} \cdot \;\frac{x}{(d^{2}+x^{2})^{1 / 2}}=m \omega ^{2} x$ $\Rightarrow \;\; \omega =(\;\frac{1}{2 \pi \epsilon _{0}} \;\frac{q^{2}}{(d^{2}+x^{2})^{3 / 2} m})^{1 / 2}$ As, $x < < d$ $\therefore \;\; \omega =(\;\frac{1}{2 \pi \epsilon _{0}} \;\frac{q^{2}}{m d^{3}})^{1 / 2}$

Given, $n=27$ $V_{1}=10 {V}$ Let $q_{1}$ be the charge of one drop, $r_{1}$ be its radius, $r$ be the radius of bigger drop and $q$ be its charge. As, volume remains constant. $\;\text{ Electric potential }\;(V)=\;\frac{3 k q_{1}^{2}}{5 r}$ $\therefore \;4 / 3 \pi r^{3}\;=4 / 3 \pi r_{1}^{3} \times 27$ $\Rightarrow \;r^{3}\;=27 r_{1}^{3}$ $\Rightarrow \;r\;=3 r_{1}$ where, $k$ is Coulomb constant. Therefore, potential energy of small drop, $U_{s}=\;\frac{3}{5} \;\frac{k q_{1}^{2}}{r_{1}}$ . . . (i) and potential energy of bigger drop, $U_{B}=\;\frac{3}{5} \;\frac{k(q_{1}^{2} \times 27)^{2}}{r}$ $U_{B}=\;\frac{3}{5} \;\frac{k(27)^{2} q_{1}^{2}}{3 r_{1}}$. . . (ii) On dividing Eq. (ii) by Eq. (i), we get $\;\frac{U_{B}}{U_{s}}=\;\frac{27 \times 27}{3}=243$ $\;U_{B}=243 U_{s}$ Hence, potential energy of big drop is 243 times of small drop.

$r=10 m m, x=2$, $|\vec{F}_{q}|=\frac{2 k \lambda}{r} \.q$ $|\vec{F}_{-q}|=\frac{2 k \lambda}{r+x} \.q$ $\Rightarrow |\vec{F}_{n e t}|=\frac{2 k \lambda q}{r}-\frac{2 k \lambda q}{r+x}$ $|\vec{F}_{n e t}|=\frac{2 k \lambdaq \.x}{r(r+x)}$ $4=\frac{2 \\times 9 \\times 10^{9} \\times 3 \\times 10^{-6} \\times q \\times 2 m m}{10 {mm} .12 {mm}}$ $\Rightarrow q=4.44 \muC$

Given, Linear charge density, $\lambda =8 \frac{nC}{m}=8 \times 10^{-9} \frac{C}{m}$ The relation between surface charge density and linear charge density can be given as $\;\frac{2 k \lambda }{r}=\;\frac{\sigma }{\epsilon _{0}}$ .....(i) where, $k=$ Coulomb's constant, $\lambda =$ linear charge density, $\sigma =$ surface charge density, $\epsilon _{0}=$ absolute electrical permittivity of free space and $r=$ distance. Substituting the values in Eq. (i), we get $\sigma =\;\frac{2 k \lambda \epsilon _{0}}{r}$ $=\;\frac{2 \times 9 \times 10^{9} \times 8 \times 10^{-9} \times 8.85 \times 10^{-12}}{3}$ $=0.424 \times 10^{-9} {Cm}^{-2}=0.424 {nCm}^{-2}$

According to Coulomb's law, the force of attraction or repulsion is directly proportional to the product of distance between them and inversely proportional to the square of the distance between them.${\text{F}}= {\text{K}} \;\frac{q_1 q_2}{r^2}$Here we have, $\text{bo} {q}_1$ is the charge of the first mass, $\text{bo} {q}_2$ is the charge of the second mass, $\text{bo} {r}$ is the distance and ${\text{K}}$ is the proportionally constant. Given: Charge of first sphere ${\text{q}}_1=2.1 {\text{nc}}$and Charge of second sphere $q_2=-0.1 {\text{NC}}$ when they connect with each other the charge becomes, $q=\;\frac{q_2-q_1}{2}$ $\Rightarrow {\text{q}}=\;\frac{2.1-0.1}{2}$ $\Rightarrow {\text{q}}=1 {\text{nC}}$ σ According to Coulomb's law, we have;${\text{F}}=\;\frac{1}{4 \pi E_o} \;\frac{q_1 q_2}{r^2}$Now, on putting the given values we have;$F=9 \times 10^9 \times \;\frac{1 \times 10^{-9} \times 1 \times 10^{-9}}{0.5^2}$ $F=9 \times 10^9 \times \;\frac{10^{-18}}{0.25}$ $\Rightarrow F=36 \times 10^{-9} {\text{N}}$ $\;\text{ Hence, }\; F=36 \times 10^{-9} {\text{N}}$

Charge of particle $\text{A}, \text{q}_{\text{A}}=20 \\mu {\text{C}}$ Charge of particle $\text{B}, \text{q}_{\text{B}}=-5 \\mu \text{C}$ Separation between A and $\text{B}, \text{R}=5 {\text{cm}}=5 \times 10^{-2}$ The given situation is shown below. Here, $\text{q}_{\text{A}}=20 \\mu \text{C}, \text{q}_{\text{B}}=-5 \\mu \text{C}, r=5 {\text{cm}}$ Let, third charge particle of charge $\text{q}_{\text{C}}$ is placed at point C to the right of charge B, so that net electric force would be equal to zero. This is because of following reason. As the net electric force on particle C should be equal to zero, so the forces due to charged particles A and B must be in opposite direction. Hence, the particle should be placed on the line AB. As particles A and B having charges of opposite signs, so charged particle C cannot be between A and B. Also, A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. Now, let position of C is at distance x from right of B. ∴ Force on C due to $\text{B}(\text{F}_{\text{CB}})=$ Force on C due to $\text{A}(\text{F}_{\text{CA}})$ $\Rightarrow {\text{k} \text{q}_{\text{B}} \text{q}_{\text{C}}}/{x^{2}}={\text{k} \text{q}_{\text{A}} \text{q}_{\text{C}}}/{(x+5 / 100)^{2}}$ $\Rightarrow \frac{(x+5 / 100)^{2}}{x^{2}}={\text{q}_{\text{A}}}/{\text{q}_{\text{B}}}$ $\Rightarrow \frac{20}{5}=(\frac{x+5 / 100}{x})^{2}$ $\Rightarrow 2=\frac{x+5 / 100}{x}$ $\Rightarrow 2 x=x+5 / 100$ ⇒ x = 5 cm

$F = ma$ $qE = ma$ $a=\frac{q E}{m}$ Now $d=\frac{1}{2} a t^{2}$ $t=\\sqrt{\frac{2 d}{a}}$ $t=\\sqrt{{2 {d}}/{(\frac{{qE}}{{m}})}}$ ${t}=\\sqrt{{2 \\times 0.1}/{(\frac{8 \\times 10^{-6}}{10^{-3}}) \\times 100}}$ $=\frac{1}{2}$ $\therefore$ Time period $=2 {t}=1 {sec}$

The given situation is shown below There can be two cases here Case 1 Considering the outer spherical shell as non-conducting.So, electric field for any point with r < R $\text{E}_{{\text{inside} }}=0$ and for $r > \text{R}, \text{E}=\frac{\text{kQ}}{r^{2}}$ In this case, option (b) is correct. Case 2 Considering the outer spherical shell as conducting. Now,charge will be induced on it.$r < \text{R}, \text{E}=0$ $\text{R} \le r < a, \text{E}=\frac{\text{kQ}}{r^{2}}$ $a \le r < b, E=0$ $r \ge b, \text{E}=\frac{\text{kQ}}{r^{2}}$ In this case, option (a) is correct.

Given, charges on two rings are +Q and -Q, respectively as shown below. Potential at centre of ring A due to charge on ring A and charge onring B is given as $\text{V}_{\text{A}}=\frac{k \text{Q}}{a}-{k \text{Q}}/{\\sqrt{s^{2}+a^{2}}}$ Similarly, potential at centre of ring B due to charge on ring B and charge on ring A is given as $\text{V}_{\text{B}}=-\frac{k \text{Q}}{a}+{k \text{Q}}/{\\sqrt{s^{2}+a^{2}}}$ Potential difference between centre of both the rings is $\Delta \text{V}=\text{V}_{\text{A}}-\text{V}_{\text{B}}$ $=\frac{k \text{Q}}{a}-{k \text{Q}}/{\\sqrt{s^{2}+a^{2}}}-(-\frac{k \text{Q}}{a}+{k \text{Q}}/{\\sqrt{s^{2}+a^{2}}})$ $=\frac{2 k \text{Q}}{a}-{2 k \text{Q}}/{\\sqrt{s^{2}+a^{2}}}$ $=\frac{\text{Q}}{2 \pi \epsilon _{0}}(\frac{1}{a}-{1}/{\\sqrt{s^{2}+a^{2}}})$ $(\because k=\frac{1}{4 \pi \epsilon _{0}} {\text{Nm}}^{2} {\text{C}}^{-2})$ Thus, potential difference between centre of both the rings is $\frac{\text{Q}}{2 \pi \epsilon _{0}}[\frac{1}{a}-{1}/{\\sqrt{s^{2}+a^{2}}}]$

We can replace $-Q$ charge at origin by $+Q$ and $-2 Q .$ Now, due to $+Q$ charge at every corner of cube, electric field at centre of cube is zero. So, net electric field at centre is only due to $-2 Q$ charge at origin. Vector form of electric field strength, $E=\;\frac{K q r}{r^{3}}$ Here, position vector, $r=\;\frac{a}{2}(\hat{x}+\hat{y}+\hat{z})$ $\Rightarrow \;\; |r|=\sqrt{(\;\frac{a}{2})^{2}+(\;\frac{a}{2})^{2}+(\;\frac{a}{2})^{2}}=\;\frac{\sqrt{3} a}{2}$ $\Rightarrow \;\; E=\;\frac{1}{4 \pi \epsilon _{0}} \times \;\frac{(-2 Q) a / 2}{(\;\frac{a}{2} \sqrt{3})^{3}} \cdot (\hat{x}+\hat{y}+\hat{z})$ E=\;{-2 Q(\hat{x}+\hat{y}+\frac\hat{z})}{3 \sqrt{3} \pi a^{2} \epsilon _{0}}

Given, charge, $q=12 \mu {C}=12 \times 10^{-6} {C}$ Height of charge from surface, $h=6 {cm}=6 \times 10^{-2} {m}$ and side of square, $a=12 {cm}=12 \times 10^{-2} {m}$ From figure, it is clear that the given square is one of the face of a cube of side $12 {cm}$ and $+12 \mu {C}$ charge is placed at its centre. Then, by Gauss's theorem,Flux through any face, $\varphi =\;\frac{q}{6 \epsilon _{0}}$ $\Rightarrow \;\; \varphi \;\; =\;\frac{12 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}}=0.226 \times 10^{6} {N}- {m}^{2} / {C}$ $\;\; =226 \times 10^{3} {N}- {m}^{2} / {C}$

Given, the side of the cube, $\text{s}=0.5 {\text{m}}$ Electric field, $\text{E}=150 \text{y}^{2} \hat{j}$ The direction of electric field is as shown in the below figure, At bottom surface, $\text{y}=0$ As we know that, the expression of electric flux, $\phi =\text{E} .\text{A} \cos \theta$ Here, E is the electric field passing through the cube and A is the surface area of the cube. Substituting the values in the above equations, we get $\phi =150 \text{y}^{2} .(0.5 \times 0.5) \times \cos180°$ $=150(0)^{2} .(0.25) \times (-1)=0$ Hence, the electric flux is zero at the bottom surface. At the top surface, $\text{y}=0.5 {\text{m}}$ Electric field, $\text{E}=150 {\text{y}}^{2} \hat{j}=150(0.5)^{2}=37.5 {\text{N}} / {\text{C}}$ Electric flux at the top surface, $\phi =\text{E} .\text{A} \cos \theta$ $=(37.5) .(0.5 \times 0.5) \cos 0°$ $=9.375 {\text{N}} / {\text{C}}-{\text{m}}^{2}$ By using the Gauss's law, $\phi ={\text{Q}_{{\in }}}/{{\epsilon _0}}$ Here, $\text{Q}_{{\in }}=$ net charge enclosed in the cube and ${\epsilon _0}=$ permittivity of the free space. Substituting the values in the above equation, we get $9.375={\text{Q}_{{\in }}}/{8.85 \times 10^{-12}}$ $\text{Q}_{{\in }}=8.3 \times 10^{-11} {\text{C}}$ The charge inside the cube is $8.3 \times 10^{-11} {\text{C}}$.

Given, length of conductor $=L$ Charge on conductor $=Q$ According to figure, $O P=a=\;\frac{\sqrt{3}}{2} L, O Q=\;\frac{L}{2}$Let $\;\; P Q=r=\sqrt{O P^{2}+O Q^{2}}$ $\Rightarrow \;\; P Q=\sqrt{(\;\frac{\sqrt{3}}{2} L)^{2}+(\;\frac{L}{2})^{2}}=\sqrt{\;\frac{3}{4} L^{2}+\;\frac{L}{4}}=L$ and $E$ be the electric field at point $P$. Since, $E$ (due to finite wire) $=\;\frac{k \lambda }{a}(\sin \varphi _{1}+\sin \varphi _{2})$. . . (i) where, $k=$ Coulomb's constant $=\;\frac{1}{4 \pi \epsilon _{0}}$ $\lambda =$ linear charge density $=\;\frac{Q}{L}$ and $\sin \varphi =\sin \varphi _{2}=\;\frac{L / 2}{L}=\;\frac{1}{2}$ Substituting the above value in Eq. (i), we get $E=\;\frac{k \lambda }{a}=\;\frac{1}{4 \pi \epsilon _{0}} \;{Q}/{L \cdot \;\frac{\sqrt{3} L}{2}}=\;\frac{1}{2 \sqrt{3} \pi \epsilon _{0} L^{2}}$

A circular disc is placed in xy-plane with its centre at the origin as shown below Consider an elemental ring of thickness dr and radius r. Now, the area of the elemental ring can be given by $\text{dA}=2 \pi \text{rdr}$ The charge stored in this elemental ring, $\text{dq}=\sigma \text{dA}$ Now, the electric field at the point on Z-axis at a distance of Z from origin can be given by $\text{dE}={\text{kdqZ}}/{(\text{r}^{2}+\text{Z}^{2})^{\frac{3}{2}}}$ Substituting the value of dq and dA in above equation, we get $\text{dE}=\frac{\text{kZ} \sigma (2 \pi \text{rdr})}{(\text{r}^{2}+\text{Z}^{2})^{3 / 2}}$ $=\frac{\sigma \text{Z}}{2 \epsilon _{0}}(\frac{\text{rdr}}{(\text{r}^{2}+\text{Z}^{2})^{3 / 2}})$ Calculating the total electric field by integrating the above expression from r = 0 to r = R, we get $\text{E}={\sigma \text{Z}}{2 \epsilon _{0}} \int _{0}^{\text{R}} \frac{\text{rdr}}{(\text{r}^{2}+\text{Z}^{2})^{3 / 2}}$ Put $r^{2}+\text{Z}^{2}=u^{2}$ $\Rightarrow 2 \text{rdr}=2 \text{udu}$ $\Rightarrow \text{rdr}=\text{ud}$ For lower limit, r = 0 ⇒ u = Z Upper limit, r = R $\Rightarrow \text{u}=\\sqrt{\text{R}^{2}+\text{Z}^{2}}$ $\text{E}=\frac{\sigma \text{Z}}{2 \epsilon _{0}} \int _{\text{Z}}^{\\sqrt{\text{R}^{2}+\text{Z}^{2}}} \frac{\text{udu}}{\text{u}^{3}}$ $\text{E}=\frac{\sigma \text{Z}}{2 \epsilon _{0}} \int _{Z}^{\\sqrt{\text{R}^{2}+\text{Z}^{2}}}\frac{d u}{u^{2}}$ $=\frac{\sigma \text{Z}}{2 \epsilon _{0}} \int _{\text{Z}}^{\\sqrt{\text{R}^{2}+\text{Z}^{2}}} u^{-2} d u$ =\frac{\sigma \text{Z}}{2 \epsilon _{0}}\[\frac{u^{-2+1}}{(-2+1)}]_{\text{Z}}^{\sqrt{\text{Z}^{2}+\text{R}^{2}}}$$=\frac{\sigma \text{Z}}{2 \epsilon {0}}[-\frac{1}{u}]{\text{Z}}^{\sqrt{\text{Z}^{2}+\text{R}^{2}}}$$=\frac{\sigma \text{Z}}{2 \epsilon _{0}}[-{1}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}+\frac{1}{\text{Z}}]$$=\frac{\sigma \text{Z}}{2 \epsilon _{0}}[\frac{1}{\text{Z}}-{1}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}]$$=\frac{\sigma }{2 \epsilon _{0}}[1-{\text{Z}}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}]$Thus, the electric field at the point on Z-axis at a distance of Z from origin is$\frac{\sigma }{2 \epsilon _{0}}[1-{\text{Z}}/{\sqrt{(\text{Z}^{2}+\text{R}^{2})}}]$

The linear charge density is $\lambda = \frac{-Q}{\text{arc length}} = \frac{-Q}{R(2\pi/3)} = \frac{-3Q}{2\pi R}$.
Consider a differential charge element $dq = \lambda R d\phi$ at an angle $\phi$ from the x-axis. The magnitude of the electric field $dE$ at O due to $dq$ is $dE = \frac{1}{4\pi\epsilon_0} \frac{|dq|}{R^2} = \frac{1}{4\pi\epsilon_0} \frac{|\lambda| R d\phi}{R^2} = \frac{|\lambda| d\phi}{4\pi\epsilon_0 R}$.
Since the charge is negative, the electric field points towards the charge element. Due to symmetry about the x-axis (from $-\pi/3$ to $\pi/3$), the y-components of the electric field will cancel out.
The x-component of the electric field is $dE_x = dE \cos\phi = \frac{|\lambda| \cos\phi d\phi}{4\pi\epsilon_0 R}$.
Integrating $E_x$ from $-\pi/3$ to $\pi/3$:
$E_x = \int_{-\pi/3}^{\pi/3} \frac{|\lambda| \cos\phi d\phi}{4\pi\epsilon_0 R} = \frac{|\lambda|}{4\pi\epsilon_0 R} [\sin\phi]_{-\pi/3}^{\pi/3}$
$E_x = \frac{|\lambda|}{4\pi\epsilon_0 R} \left(\sin(\pi/3) - \sin(-\pi/3)\right) = \frac{|\lambda|}{4\pi\epsilon_0 R} \left(\frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right)\right) = \frac{|\lambda|\sqrt{3}}{4\pi\epsilon_0 R}$
Substitute $|\lambda| = \frac{3Q}{2\pi R}$:
$E_x = \frac{3Q}{2\pi R} \frac{\sqrt{3}}{4\pi\epsilon_0 R} = \frac{3\sqrt{3} Q}{8\pi^2\epsilon_0 R^2}$
The total electric field is in the positive x-direction: $\vec{E} = \frac{3\sqrt{3} Q}{8\pi^2\epsilon_0 R^2} \hat{i}$.

The final answer is $\boxed{\text{D}}$.

Given, number of mercury drops, $n=512$ Voltage of each drop, $V=2 V$ Let $r, R$ be the radius of drop small and combined spherical drop, respectively. Now, when all drops are joined into single drop, volume remains constant, i.e. $\;\; 512 \times \;\frac{4}{3} \pi r^{3}=\;\frac{4}{3} \pi R^{3}$ $\because \;\; V=\;\frac{k q}{r}=2$. . . (i) $\therefore \;\; V_{\;\text{net }\;}=\;\frac{k Q}{R}$ where, $Q$ be the charge of bigger sphere. $\Rightarrow \;\; V_{\;\text{net }\;}=\;\frac{k \times 512 q}{8 r}$ . . . (ii) On dividing Eq. (ii) by Eq. (i), we get $\;{V_{\;\text{net }\;}}/{V}=\;\frac{k \times 512 q \times r}{8 r \times k q}=\;\frac{512}{8}$ $\;V_{\;\text{net }\;}=2 \times \;\frac{512}{8}=128 {V}$

Given, {E}=\;\frac{3 E_{0}}{5} { {i}^{∧}+\;\frac{4}{5} E_{0} { {j}^{∧}, $A_{1}=0.2 {m}^{2} \hat{i}$ and $A_{2}=0.3 {m}^{2} \hat{j}$ Let $\varphi _{1}$ and $\varphi _{2}$ be the flux linked with area $A_{1}$ and $A_{2}$, respectively. As we know that, $\varphi \;\; =∮ E \cdot d S=E \cdot A$ $\Rightarrow \;\; \varphi _{1} \;\; =(3 / 5 E_{0} \hat{i}+4 / 5 E_{0} \hat{j}) \cdot 0.2 \hat{i}=3 / 5 E_{0} \times 0.2$ and similarly, $\;\; \varphi _{2}=4 / 5 E_{0} \times 0.3$ Now, $\;\; \;\frac{\varphi _{1}}{\varphi _{2}}=\;\frac{3 / 5 E_{0} \times 0.2}{4 / 5 E_{0} \times 0.3}=\;\frac{0.6}{1.2}=\;\frac{1}{2}$ $\therefore \;\; a=1$