Electrostatics

Given data:

• Dipole moment, $ p = 2 \times 10^{-10}\ \mathrm{C \, m} $

• Electric field, $ E = 10^4\ \mathrm{N \, C^{-1}} $

• Angle between $ \vec{p} $ and $ \vec{E} $, $ \theta = 30^\circ $

Formula for torque:

$ \tau = p E \sin \theta $

Substitute values:

$ \tau = (2 \times 10^{-10})(10^4) \sin 30^\circ $

$ = 2 \times 10^{-6} \times \frac{1}{2} $

$ = 1 \times 10^{-6}\ \mathrm{N \, m} $

✅ Final Answer:

$ \boxed{10^{-6}\ \mathrm{N \, m}} $

Correct option: A

$ \begin{aligned} &\begin{aligned} O A & =\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2 \\ O B & =2 \mathrm{~m} \\ \therefore \quad V_A & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{O A} \\ & =9 \times 10^9 \times \frac{10^{-3} \times 10^{-6}}{2}=4.5 \mathrm{~V} \end{aligned}\\ &\text { Similarly, } V_B=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{O B}\\ &\begin{aligned} & =9 \times 10^9 \times \frac{10^{-3} \times 10^{-6}}{2} \\ & =4.5 \mathrm{~V} \\ \therefore \quad V_A-V_B & =4.5-4.5=0 \mathrm{~V} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_A q_B}{r^2} \\ = & 9 \times 10^9 \times \frac{1.6 \times 10^{-19} \times 3.2 \times 10^{-19}}{9 \times 10^{-4}} \\ = & 512 \times 10^{-25} \mathrm{~N} \end{aligned} $

$ \begin{aligned} & \text { Diagonal } A C=\sqrt{A D^2+D C^2} \\ & =\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2 \mathrm{~m} \\ & \therefore O A=O B=O C=O D=\frac{A C}{2}=\frac{2}{2}=1 \mathrm{~m} \end{aligned} $

Electrostatic potential of the centre $(O)$

$ \begin{aligned} V= & V_A+V_B+V_C+V_D \\ = & \frac{1}{4 \pi \varepsilon_0} \cdot \frac{12 \times 10^{-9}}{1}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(-20 \times 10^{-9}\right)}{1} +\frac{1}{4 \pi \varepsilon_0} \frac{32 \times 10^{-9}}{1}+\frac{1}{4 \pi \varepsilon_0} \frac{\left(-15 \times 10^{-9}\right)}{1} \\ = & \frac{10^{-9}}{4 \pi \varepsilon_0}[12-20+32-15] \\ = & 9 \times 10^9 \times 10^{-9}(9)=81 \text { volt } \end{aligned} $

$ \begin{aligned} &\text { Initially, }\\ &F=\frac{k\left|q_1 q_2\right|}{r^2} \end{aligned} $

$ \begin{aligned} & =\frac{k \times 8 \times 10^{-6} \times 4 \times 10^{-6}}{r^2} \\ & =\frac{32 \times 10^{-12} k}{r^2} \end{aligned} $

After connecting by conducting wires,

$ q=q_1+q_2=8 \mu C+(-4 \mu C)=4 \mu C $

∴ Charge on each sphere

$ \begin{aligned} q^{\prime} & =\frac{q}{2}=2 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C} \\ \therefore \quad F^{\prime} & =k \cdot \frac{q^2}{r^2}=\frac{k \times\left(2 \times 10^{-6}\right)^2}{r^2} \\ & =\frac{4 \times 10^{-12} k}{r^2}=\frac{1}{8} \times \frac{32 \times 10^{-12} k}{r^2} \\ & =\frac{F}{8} \end{aligned} $

$ \begin{aligned} & \text { } V=20|r|=20 \sqrt{x^2+y^2+z^2} \\ & \begin{aligned} & \therefore E=-\left(\frac{\partial v}{\partial x} \hat{\mathbf{i}}+\frac{\partial v}{\partial y} \hat{\mathbf{j}}+\frac{\partial v}{\partial z} \hat{\mathbf{k}}\right) \\ &=-\left[\frac{20 x}{\sqrt{x^2+y^2+z^2}} \hat{\mathbf{i}}+\frac{20 y}{\sqrt{x^2+y^2+z^2}} \hat{\mathbf{j}}\right. \left.+\frac{20 z}{\sqrt{x^2+y^2+z^2}} \hat{\mathbf{k}}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{-20(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{\sqrt{x^2+y^2+z^2}} \\ & \text { At point }(4 m, 3 m,-5 m) \\ & \qquad \begin{aligned} E & =\frac{-20(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})}{\sqrt{4^2+3^2+(-5)^2}} \\ & =\frac{-20}{5 \sqrt{2}}(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}) \\ & =-\sqrt{2}(8 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}) \end{aligned} \end{aligned} $

Given, $q_1+q_2=25 \mu \mathrm{C}$

Also, $F=\frac{k q_1 q_2}{r^2}$

$ \begin{aligned} & \Rightarrow \quad 0.6=\frac{9 \times 10^9\left(q_1 q_2\right)}{(1.5)^2} \\ & \Rightarrow \quad q_1 q_2=\frac{0.6 \times(1.5)^2}{9 \times 10^9} \\ & \Rightarrow q_1 q_2=1.5 \times 10^{-10} \end{aligned} $

Using $\left(q_1-q_2\right)^2=\left(q_1+q_2\right)^2-4 q_1 q_2$

we get

$ \begin{aligned} & \left(q_1-q_2\right)^2=\left(25 \times 10^{-6}\right)^2-4 \times 1.5 \times 10^{-10} \\ & \Rightarrow\left(q_1-q_2\right)^2=625 \times 10^{-12}-6 \times 10^{-10} \\ & =(625-600) \times 10^{-12} \\ & =25 \times 10^{-12} \end{aligned} $

or $\quad q_1-q_2=5 \times 10^{-6} \mathrm{C}=5 \mu \mathrm{C}$

Given:

• Total atoms in the solid = $6 \times 10^{24}$

• $0.005\%$ of atoms lose one electron each.

Step 1: Find how many atoms lose an electron

$ \text{Number of affected atoms} = (0.005\% ) \times (6 \times 10^{24}) $

$ 0.005\% = \frac{0.005}{100} = 5\times10^{-5} $

$ \Rightarrow \text{Number of atoms losing electrons} = 6 \times 10^{24} \times 5\times10^{-5} = 3 \times 10^{20} $

Step 2: Each atom loses one electron

Each lost electron corresponds to a positive charge gained by the solid equal to the charge of that electron ($e = 1.6\times10^{-19}\ \text{C}$).

Step 3: Total charge gained

$ Q = (3\times10^{20}) \times (1.6\times10^{-19}) $

$ Q = 4.8\times10^{1} = 48\ \text{C} $

✅ Final Answer:

$ \boxed{+48\ \text{C}} $

Correct Option: B (+48 C)

Step 1: Write the Formula for Energy Stored

The energy stored in a capacitor is given by the formula: $U = \frac{q^2}{2C}$ where $U$ is energy, $q$ is charge, and $C$ is capacitance.

Step 2: Rearranging to Find Capacitance

We can solve for $C$ as follows: $C = \frac{q^2}{2U}$

Here, $q = 12 \times 10^{-6}~\text{C}$ and $U = 6~\text{J}$. Substitute these values into the formula:

$C = \frac{(12 \times 10^{-6})^2}{2 \times 6}$

Step 3: Calculate Capacitance

Simplify the above equation: $C = 12 \times 10^{-12}~\text{F}$

Step 4: Capacitance of a Spherical Conductor

The capacitance of a spherical conductor is given by: $C = 4\pi\varepsilon_0 r$

Step 5: Find the Radius

Rearrange to find the radius $r$: $r = \frac{C}{4\pi\varepsilon_0}$

We can also use $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9~\text{N m}^2/\text{C}^2$ so:

$r = 9 \times 10^9 \times 12 \times 10^{-12}$

Step 6: Solve for the Radius

Multiply the values: $r = 108 \times 10^{-3} = 0.108~\text{m} = 10.8~\text{cm}$

Force on charge particle in electric field,

$ F=q E=2 \times 20=40 \mathrm{~N} $

According to work energy theorem, Kinetic energy $=W$

$ =F \cdot s=40 \times 0.2=8 \mathrm{~J} $

$\sigma_1=\sigma_2=\sigma$

Since,  $ E=\frac{\sigma}{\varepsilon_0} $

$\therefore$ For the smaller sphere,

$ E_1=\frac{\sigma_1}{\varepsilon_0}=\frac{\sigma}{\varepsilon_0} $

For the larger sphere,

$ E_2=\frac{\sigma_2}{\varepsilon_0}=\frac{\sigma}{\varepsilon_0} $

Since, $\quad \sigma_1=\sigma_2=\sigma$

Thus, $\quad E_1=E_2=E$

As we move along a field line we are moving in direction of reducing potential.

So,  $ V_C>V_B>V_A $

Potential at point $C$, due to charges $+q$ and $-q$.

$ V_C=-\frac{k q}{d}+\frac{k q}{d}=0 $

and potential of point $D$

$ V_0=\frac{-k q}{d}+\frac{k q}{3 d}=\frac{-2 k q}{3 d} $

So, potential difference between $C$ and $D$

$ V_D-V_C=\frac{-2 k q}{3 d} $

So, work done in moving charge $Q$ from $C$ to $D$

$ \begin{aligned} & W=Q\left(V_D-V_C\right) \\ & W=\frac{-2 k Q q}{3 d} \end{aligned} $

$ \begin{aligned}Now,\,\, & K=\frac{1}{4 \pi \varepsilon_0} \Rightarrow W=\frac{-2 Q q}{3\left(4 \pi \varepsilon_0\right) d} \\ & W=\frac{-Q q}{6 \pi \varepsilon_0 d} \end{aligned} $

$ \begin{aligned} &\text { Electric flux }\\ &\begin{aligned} \phi & =\mathbf{E} \cdot \mathbf{A}=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \cdot(3 \hat{\mathbf{i}}) \\ & =9\left(\mathrm{Nm}^2 / \mathrm{C}\right) \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \text { } & \frac{1}{2} m v^2=q V \\ \Rightarrow \quad v & =\sqrt{\frac{2 q V}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 180}{9 \times 10^{-31}}} \\ & =8 \times 10^6 \mathrm{~m} / \mathrm{s}=8000 \times 10^3 \mathrm{~m} / \mathrm{s} \\ & =8000 \mathrm{~km} / \mathrm{s} \end{aligned} $

Initial potential energy of the system of

$ U_i=3 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} $

Final potential energy of the system

$ U_f=3 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{\frac{L}{2}}=6 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} $

$\therefore$ Required work done

$ \begin{aligned} W & =U_f-U_i \\ & =6 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L}-3 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{3 q^2}{L} \end{aligned} $

$27 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3$

$ \Rightarrow \quad R=3 r $

Potential on small drop

$ V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}=9 \times 10^9 \times \frac{10^{-12}}{10^{-3}}=9 \mathrm{volt} $

Potential on big drop

$ =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{27 q}{R}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{27 q}{3 r} $

$ \begin{aligned} & =9 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ & =9 \times 9=81 \mathrm{~V} \end{aligned} $

$F_0=98 \mathrm{~N}$

$ \begin{aligned} & \text { Using, } \frac{d}{k}=\frac{d_1}{k_1}+\frac{d_2}{k_2} \\ & \Rightarrow \quad \frac{9}{k}=\frac{6}{4}+\frac{3}{9} \\ & \Rightarrow \quad k=\frac{54}{11} \end{aligned} $

$\therefore$ New force, $F^{\prime}=\frac{F_0}{k}$

$ =\frac{98}{\frac{54}{11}}=98 \times \frac{11}{54}=19.96 \mathrm{~N} $

Which is closest to 18 N .

$U_{\text {initial }}=U_{12}+U_{13}+U_{23}$

$ =\frac{k q^2}{2 a}-\frac{k q^2}{a}-\frac{k q^2}{a}=\frac{k q^2}{2 a}-\frac{2 k q^2}{a}=\frac{-3 k q^2}{2 a} $

After exchanging the final charge with one of the negative charge, there are two - $q$ charges separated by a distance $a$ and one $+q$ charge separated by a distance ' $a$ ' from one of the $-q$ charges.

$ \begin{aligned} & \therefore U_{\text {final }}=U_{12}^{\prime}+U_{13}^{\prime}+U_{23}^{\prime} \\ & =\frac{k q^2}{a}-\frac{k q^2}{2 a}-\frac{k q^2}{a}=\frac{-k q^2}{2 a} \\ & \begin{aligned} \therefore \Delta U & =U_{\text {final }}-U_{\text {initial }} \\ & =-\frac{k q^2}{2 a}-\left(\frac{-3 k q^2}{2 a}\right)=\frac{2 k q^2}{2 a} \\ & =\frac{k q^2}{a}=\frac{q^2}{4 \pi \varepsilon_0 a} \quad\left[\because k=\frac{1}{4 \pi \varepsilon_0}\right] \end{aligned} \end{aligned} $

Mass, $ m = 0.5 \, \text{g} = 0.5 \times 10^{-3} \, \text{kg} $

Charge, $ q = 10 \times 10^{-6} \, \text{C} $

Electric field, $ E = 8 \, \text{NC}^{-1} $

First, calculate the force $ F $ on the particle using the equation:

$ F = q \times E $

Substituting the values:

$ F = 10 \times 10^{-6} \times 8 = 80 \times 10^{-6} = 8 \times 10^{-5} \, \text{N} $

Next, find the acceleration $ a $ using the formula:

$ a = \frac{F}{m} $

Substitute the values to find:

$ a = \frac{8 \times 10^{-5}}{0.5 \times 10^{-3}} = 0.16 \, \text{m/s}^2 $

Since the particle starts from rest, the initial velocity $ u $ is $ 0 $.

Using the kinematic equation for velocity:

$ v = u + at $

Substituting the known values:

$ v = 0 + 0.16 \times 5 = 0.8 \, \text{m/s} $

Therefore, the velocity of the particle after 5 seconds is $ 0.8 \, \text{m/s} $.

The electric potential $V$ of a sphere is directly proportional to its charge $Q$ and inversely to its radius $r$,

$ V=\frac{k Q}{r} $

For small sphere,

$ \begin{aligned} V_{\text {small }} & =\frac{k Q_{\text {small }}}{r_{\text {small }}}=60 \mathrm{mV} \\ V_{\text {big }} & =\frac{k Q_{\text {total }}}{r_{\text {big }}}=\frac{k \cdot 125 Q_{\text {small }}}{r_{\text {big }}} \end{aligned} $

The volume of the big sphere is 125 times the volume of small sphere, so

$ \begin{aligned} V_{\text {big }} & =\frac{k \times 125 Q_{\text {small }}}{\left(125^{1 / 3}\right) r_{\text {small }}} \\ & =125^{2 / 3} \cdot \frac{k Q_{\text {small }}}{r_{\text {small }}} \\ & =125^{2 / 3} \cdot 60 \mathrm{mV} \\ & =125^{2 / 3} \times 60 \times 10^{-3}=1.5 \mathrm{~V} \end{aligned} $

Therefore, the electric potential on the bigger formed is 1.5 V .

$ \begin{aligned} & \text { Given, } q_1=4 \mathrm{nC}=4 \times 10^{-9} \mathrm{C} \\ & U=1.8 \mu \mathrm{~J}=1.8 \times 10^{-6} \mathrm{~J} \\ & r=10 \mathrm{~cm}=0.1 \mathrm{~m} \end{aligned} $

$ \begin{aligned} & \text { As, } Q=\frac{U \times r}{k \times q_1} \\ & \Rightarrow Q=\frac{1.8 \times 10^{-6} \times 0.1}{8.9875 \times 10^9 \times 4 \times 10^{-9}} \\ & \Rightarrow Q=\frac{1.8 \times 10^{-7}}{35.95} \\ & \Rightarrow Q=5 \times 10^{-9} \mathrm{C} \\ & \Rightarrow Q=5 \mathrm{nC} \end{aligned} $

Given:

Mass of a deuteron, $ m_d = 3.2 \times 10^{-27} \, \text{kg} $

Charge of a deuteron, $ q_d = 1.6 \times 10^{-19} \, \text{C} $

Gravitational acceleration, $ g = 9.8 \, \text{m/s}^2 $

To suspend the deuteron in the air, the electric force ($ F_\text{electric} $) must equal the gravitational force ($ F_\text{gravity} $):

$ F_\text{electric} = F_\text{gravity} $

Thus:

$ qE = mg $

Solving for $ E $, the magnitude of the electric field:

$ E = \frac{mg}{q} = \frac{3.2 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}} $

$ E = \frac{3.2 \times 9.8}{1.6} \times 10^{-8} $

$ E = 2 \times 9.8 \times 10^{-8} = 19.6 \times 10^{-8} \, \text{N/C} $

Therefore, the magnitude of the electric field required is $ 19.6 \times 10^{-8} \, \text{N/C} $.

Given:

$ q_1 = 5 \, \text{nC} = 5 \times 10^{-9} \, \text{C} $

$ q_2 = -2 \, \text{nC} = -2 \times 10^{-9} \, \text{C} $

These charges are positioned along the X-axis. First, let's determine the distance between them:

$ \begin{align*} x &= x_2 - x_1 \\ &= (23 - 5) \, \text{cm} = 18 \, \text{cm} = 18 \times 10^{-2} \, \text{m} \end{align*} $

Now, calculate the electrostatic potential energy ($U$) of the charges using the formula:

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{x} $

Substitute the known values:

$ U = \frac{9 \times 10^9 \times 5 \times 10^{-9} \times (-2 \times 10^{-9})}{18 \times 10^{-2}} $

$ U = -5 \times 10^{-7} \, \text{J} $

The negative sign in the electrostatic potential energy indicates that the force between the two charges is attractive.

Given:

Electric dipole moment, $ p = 5 \times 10^{-30} \, \text{C} \cdot \text{m} $

In a neutral ammonia (NH$_3$) molecule, there are 10 electrons and 10 protons. The electric dipole moment is expressed as:

$ p = q \times 2l = 10e \times 2l $

Where $ e $ is the elementary charge ($1.6 \times 10^{-19} \, \text{C}$).

To find the separation distance $ 2l $, we use:

$ 2l = \frac{p}{10e} = \frac{5 \times 10^{-30}}{10 \times 1.6 \times 10^{-19}} $

Calculating this gives:

$ 2l = 3.125 \times 10^{-12} \, \text{m} $

Thus, the distance between the centers of positive and negative charge in the molecule is $ 3.125 \times 10^{-12} \, \text{m} $.

$ \begin{aligned} & \text { } q_1=+1 \times 10^{-8} \mathrm{C} \\ & q_2=-2 \times 10^{-8} \mathrm{C} \\ & q_3=+3 \times 10^{-8} \mathrm{C} \\ & q_4=+2 \times 10^{-8} \mathrm{C} \end{aligned} $

Here,

At point $A$ and $C$, charges are $q_1=1 \times 10^{-8} \mathrm{C}$ and $3 \times 10^{-1} \mathrm{C}$

$ \begin{aligned} V_{A C} & =9 \times 10^9\left[\frac{1 \times 10^{-8}+3 \times 10^{-8}}{1}\right] \sqrt{2} \\ & =9 \times 10^9\left[4 \times 10^{-8}\right] \sqrt{2} \end{aligned} $

$ \begin{aligned} & =9 \times 10^9 \times 4 \times 10^{-8} \times \sqrt{2} \\ & =90 \times 4 \times \sqrt{2}=360 \sqrt{2}=510 \mathrm{~V} \end{aligned} $

At point $B$ and $D$, charges are $q_2=-2 \times 10^{-8} \mathrm{C}$ and $q_4=2 \times 10^{-8} \mathrm{C}$ Thus potential $V_{B D}=0$ Hence, answer will be 510 V

Given:

Distance $ r = 3 \, \text{m} $

Linear charge density $ \lambda = 0.2 \, \mu\text{C/m} = 0.2 \times 10^{-6} \, \text{C/m} $

The electric field $ E $ at a distance $ r $ from a long straight wire is calculated using the formula:

$ E = \frac{\lambda}{2\pi \varepsilon_0 r} $

Where:

$\varepsilon_0$ is the permittivity of free space, approximately $8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$

Substitute the given values into the formula:

$ E = \frac{0.2 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12} \times 3} $

This calculation results in:

$ E \approx 1.2 \times 10^3 \, \text{V/m} $

Given:

A charge $ q $ is placed at the center of a cube.

The side length of the cube is $ L $.

We know that the electric flux through a closed surface that encloses a charge $ q $ is given by Gauss’s Law:

$ \phi_t = \frac{q}{\varepsilon_0} $

Here, $\phi_t$ represents the total electric flux through the entire surface of the cube.

Since a cube has 6 faces, the flux through each individual face, $\phi_f$, is the total flux divided equally among the six faces:

$ \phi_f = \frac{1}{6} \times \phi_t = \frac{1}{6} \times \frac{q}{\varepsilon_0} = \frac{q}{6 \varepsilon_0} $

This shows that the electric flux associated with each face of the cube is $\frac{q}{6 \varepsilon_0}$.

Given:

Each side of the equilateral triangle has a length $L$.

Each vertex has a charge $q$.

The potential energy $U$ of the system can be calculated by considering the interactions between each pair of charges. Since there are three sides in an equilateral triangle, we have three pairs of charges to consider:

$ U = \frac{1}{4 \pi \varepsilon_0}\left[\frac{q \cdot q}{L} + \frac{q \cdot q}{L} + \frac{q \cdot q}{L}\right] $

Simplifying the expression:

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} \cdot (1 + 1 + 1) $

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{3q^2}{L} $

Therefore, the potential energy of the system of three equal charges at the vertices of an equilateral triangle is:

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{3q^2}{L} $

To understand the change in capacitance when eight smaller drops combine to form a larger drop, consider the following:

Let $ R $ be the radius of the larger drop and $ r $ be the radius of each smaller drop. When these smaller drops combine, the total volume remains constant. Therefore, the volume of the large drop equals the total volume of the eight smaller drops:

$ \frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3 $

This simplifies to:

$ R^3 = 8r^3 $

$ R = 2r $

Given that the capacitance $ C $ of an isolated spherical conductor is described by the formula:

$ C = 4 \pi \varepsilon_0 r $

The capacitance of each smaller drop is:

$ C = 4 \pi \varepsilon_0 r $

For the larger drop, now having a radius $ R = 2r $, the capacitance $ C' $ is calculated as follows:

$ C' = 4 \pi \varepsilon_0 R $

$ C' = 4 \pi \varepsilon_0 (2r) = 2 \times 4 \pi \varepsilon_0 r = 2C $

Thus, the capacitance of the larger drop is $ 2C $.

Electrostatic force on $+q$ due to $+2 q$ and $+4 q$

$E_{q, 2 q}=\frac{k q(2 q)}{d^2}=\frac{2 k q^2}{d^2} \quad \ldots \text { (i) } $

$E_{q, 4 q}=\frac{k q(4 q)}{(2 d)^2}=\frac{k q^2}{d^2} \quad \ldots \text { (ii) } $

Total force on $+q$

$ =\frac{2 k q^2}{d^2}+\frac{k q^2}{d^2}=\frac{3 k q^2}{d^2} \quad \ldots \text { (iii) } $

Electrostatic force on $+4 q$ due to $+2 q$ and $+q$

$E_{4 q, 2 q}=\frac{k(2 q)(4 q)}{d^2}=\frac{8 k^2}{d^2} \quad \ldots \text { (iv) } $

$E_{4 q, q}=\frac{k(4 q)(q)}{(2 d)^2}=\frac{k q^2}{d^2} \quad \ldots \text { (v) } $

Total force on $+4 q=9 \frac{k q^2}{d^2} \quad \ldots \text { (vi) } $

Ratio of net force on $q_1$ and $+4 q$.

$ =\frac{3 k q^2}{d^2} \times \frac{d^2}{9 k q^2}=\frac{3}{9}=1: 3 $

To determine the speed acquired by a particle when it is accelerated through a potential difference, we need to use the concept of energy conservation. The kinetic energy gained by the particle is equal to the electrical work done on the particle due to the potential difference.

Given:

Mass of the particle, $m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg}$

Charge of the particle, $Q = 6 \, \mu\text{C} = 6 \times 10^{-6} \, \text{C}$

Potential difference, $V = 60 \, \text{V}$

The relationship between the kinetic energy (KE) gained and the potential difference is described by the equation:

$ \text{KE} = Q \times V $

The kinetic energy can also be expressed in terms of velocity ($v$) as:

$ \frac{1}{2} m v^2 = Q V $

Solving for velocity $v$, we get:

$ v = \left(\frac{2 Q V}{m}\right)^{1/2} $

Substituting the given values into the equation:

$ v = \left(\frac{2 \times 6 \times 10^{-6} \times 60}{2 \times 10^{-3}}\right)^{1/2} $

Simplifying further:

$ v = \left(\frac{720 \times 10^{-6}}{2 \times 10^{-3}}\right)^{1/2} $

$ v = \left(0.36\right)^{1/2} $

$ v = 0.6 \, \text{m/s} $

Thus, the speed acquired by the particle is $0.6 \, \text{m/s}$.

When the two spheres $A$ and $B$ are connected by a fine wire, then the electric potential will be same on bedt the spheres.

$ \begin{aligned} & \Rightarrow \quad V_A=V_B \Rightarrow \frac{k q_1}{r_1}=\frac{k q_2}{r_2} \\ & \Rightarrow \quad \frac{q_1}{0.4}=\frac{q_2}{0.6} \Rightarrow q_2=1.5 q_1 \end{aligned} $

Total charges is conserved here,

Charge before connecting wire $=\mathrm{charge}$ after connecting wire.

$ \begin{aligned} 80 \mu \mathrm{C}+40 \mu \mathrm{C} & =2.5 q_1 \\ q_1=\frac{120 \mu \mathrm{C}}{2.5} & =48 \mu \mathrm{C} \end{aligned} $

and $\quad q_2=1.5 q_1=1.5 \times 48=72 \mu \mathrm{C}$ So, we can conclude that $(72-40) \mu \mathrm{C}=32$ $\mu C$ charge flows from sphere $A$ to $B$.

From the above figure, it can be concluded that the dipole moment of a dipole and electric field strength are opposite to each other on equatorial line. Thus, the angle between them is $180^{\circ}$.

In the given scenario, two particles with equal mass $ m $ and charge $ q $, separated by a distance of 16 cm, are in a state where they do not experience any net force. This implies that the electrostatic force and gravitational force between the particles are equal and opposite.

Initially, the electrostatic force ($ F_c $) between the particles is given by Coulomb's Law:

$ F_c = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} $

where $ \varepsilon_0 $ is the permittivity of free space, $ q $ is the charge, and $ r $ is the separation distance.

The gravitational force ($ F_g $) can be expressed using Newton's Law of Gravitation:

$ F_g = \frac{G m^2}{r^2} $

where $ G $ is the universal gravitational constant and $ m $ is the mass of each particle.

According to the problem, $ F_c = F_g $. Therefore, we equate the two equations:

$ \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} = \frac{G m^2}{r^2} $

Canceling $ r^2 $ from both sides:

$ \frac{q^2}{4 \pi \varepsilon_0} = G m^2 $

This simplifies to:

$ \frac{q^2}{m^2} = 4 \pi \varepsilon_0 G $

Taking the square root of both sides gives us:

$ \frac{q}{m} = \sqrt{4 \pi \varepsilon_0 G} $

This expression gives the value of $\frac{q}{m}$ in terms of known constants.

For the given situation, points $A, B$ and $C$ are at equipotential surface.

$ \begin{array}{ll} \text { i.e., } & V_A=V_B=V_C \\ \therefore & W_A=W_B=W_C=0 \end{array} $

Given:

Initial charges:

$ q_1 = +6 \, \mu \mathrm{C} = 6 \times 10^{-6} \, \mathrm{C} $

$ q_2 = +10 \, \mu \mathrm{C} = 10 \times 10^{-6} \, \mathrm{C} $

Force of repulsion: $ F = 30 \, \mathrm{N} $

Using Coulomb's Law:

$ F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} $

Solve for $ r^2 $:

$ r^2 = \frac{9 \times 10^9 \times 6 \times 10 \times 10^{-12}}{30} $

$ r^2 = 18 \times 10^{-3} \quad \ldots \text{(Equation 1)} $

Adding Additional Charges:

Each charge receives an additional $-8 \mu \mathrm{C}$:

$ q_1' = +6 - 8 = -2 \mu \mathrm{C} $

$ q_2' = +10 - 8 = +2 \mu \mathrm{C} $

Recalculate the Force Using Coulomb's Law:

$ F' = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1' \cdot q_2'}{r^2} $

Substitute $ q_1' $ and $ q_2' $:

$ F' = \frac{9 \times 10^9 \times 2 \times 2 \times 10^{-12}}{18 \times 10^{-3}} $

(From Equation 1)

$ F' = 2 \, \mathrm{N} $

Since one charge is positive and the other is negative, they attract each other with a force of $ 2 \, \mathrm{N} $.

To find the electric potential at the point $(1 \, \text{m}, 2 \, \text{m})$, we begin with the given electric field and displacement vector:

$ \mathbf{E} = (30 \hat{\mathbf{i}} + 40 \hat{\mathbf{j}}) \, \text{N/C} $

$ \mathbf{d} = (1 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}) \, \text{m} $

The formula for electric potential $V$ with respect to a chosen reference point is given by:

$ V = -\mathbf{E} \cdot \mathbf{r} $

Applying the formula, we compute the dot product:

$ V = -(30 \hat{\mathbf{i}} + 40 \hat{\mathbf{j}}) \cdot (1 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}) $

Break it down as follows:

$ V = -(30 \cdot 1 + 40 \cdot 2) $

$ V = -(30 + 80) $

$ V = -110 \, \text{V} $

Thus, the electric potential at the point $(1 \, \text{m}, 2 \, \text{m})$ is $-110 \, \text{V}$.

The given situation is shown below

From the above figure total charge enclosed by the sphere of radius $2.5 \mathrm{~cm}=5 \mathrm{q}$

$\therefore$ According to gauss law, total electric flux passing through spherical surface,

$\begin{aligned} \phi & =\frac{\text { total charge enclosed }}{\varepsilon_0} \\ & =\frac{5 q}{\varepsilon_0} \end{aligned}$

The given situation is shown below

Since, electric field is directed along the $X$-axis, hence, point $A$ and $C$ lie on equipotential surface

$\therefore$ Potential at point $C=$ Potential at point $A$

$\begin{array}{rlrl} \Rightarrow & V_C =V_A \\ \therefore & V_B-V_A =V_B-V_C \\ & =-E \Delta x=-5(2-0)=-10 \mathrm{~V} \end{array}$

According to given situation, electric field produced in the region of drilled hole at a distance $x$ from its centre,

$E=\frac{Q x}{4 \pi \varepsilon_0 R^3}\quad \text{.... (i)}$

Force on charge $(-q)$ when it is placed at a distance $x$,

$\begin{aligned} & F=-q E \\ & \Rightarrow \quad m a=-q E \\ & a=\frac{-q E}{m}=\frac{-q}{m} \cdot \frac{Q x}{4 \pi \varepsilon_0 R^3} \\ & \Rightarrow \quad a=\frac{-Q q x}{4 \pi \varepsilon_0 m R^3} \\ & \Rightarrow \quad-\omega^2 x=\frac{-Q q x}{4 \pi \varepsilon_0 m R^3} \\ & \therefore \quad \omega^2=\frac{Q q}{4 \pi \varepsilon_0 m R^3} \\ & (2 \pi f)^2=\frac{Q q}{4 \pi \varepsilon_0 m R^3} \\ & \Rightarrow \quad f=\frac{1}{2 \pi}\left(\frac{Q q}{4 \pi \varepsilon_0 m R^3}\right)^{1 / 2} \\ & \end{aligned}$

We know that, electric field $(E)$ and electric potential are related as

$E=\frac{-d V}{d r}$

For the region of constant potential $(V)$

$E=0$

According to Gauss's, law, electric flux through closed Gaussian surface

$\phi=\oint E . d s=\frac{q}{\varepsilon_0}\quad \text{... (i)}$

$\begin{aligned} & \text { if } \quad E=0 \\ & \Rightarrow \quad \oint E . d s=0 \\ \end{aligned}$

Hence, from Eq (i) , $q=0$

Inside a charged hollow metal sphere, electric field $(E)$ is zero but electric potential $(V)$ is not zero and equal to same potential as on its surface. Work done to move a charge particle (either $+v e$ or $-v e$) on a equipotential surface between any two points is always zero.

$\begin{aligned} \text { Since, } W & =q\left(V_1-V_2\right) \\ & =q(V-V) \quad \left(\because V_1=V_2=V\right)\\ & =0 \end{aligned}$

Potential energy of system of two like charges placed at a distance $r$ is given as,

$U=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}$

When both like charges $q_1$ and $q_2$ are brought closer, then $r$ decreases, hence potential energy $U$ increases.

Therefore, statements (A), (B) and (C), all are true.

According to first condition,

$F=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \quad \text{... (i)}$

Electric field produced due to dielectric material,

$\begin{aligned} E & =\frac{9}{4 \pi \varepsilon_0\left(r-\frac{r}{5}+\sqrt{k} \frac{r}{5}\right)^2} \\ & =\frac{q}{4 \pi \varepsilon_0 \frac{r^2}{25}(4+\sqrt{k})^2} \\ & =\frac{q}{4 \pi \varepsilon_0 \frac{r^2}{25}}(4+\sqrt{9})^2 \quad (\because k=9)\\ E & =\frac{q}{4 \pi \varepsilon_0 r^2}\left(\frac{25}{49}\right) \end{aligned}$

Force between the charges,

$\begin{aligned} F^{\prime} & =q E=q \cdot \frac{q}{4 \pi \varepsilon_0 r^2}\left(\frac{25}{49}\right) \\ & =\frac{25}{49} \frac{q^2}{4 \pi \varepsilon_0 r^2} \\ & =\frac{25}{49} F \end{aligned}$

Electric dipole moment,

$p=5 \times 10^{-7} \mathrm{C}$

Angle, $\theta=60^{\circ}$

Electric field, $E=2 \times 10^4 \mathrm{~N} / \mathrm{C}$

Torque,

$\begin{aligned} \tau & =p E \sin \theta \\ & =5 \times 10^{-7} \times 2 \times 10^4 \times \sin 60^{\circ} \\ & =10^{-2} \times \frac{\sqrt{3}}{2}=8.66 \times 10^{-3} \mathrm{~N}-\mathrm{m} \end{aligned}$

$\begin{aligned} & \text { Given, } q_1=10 \mu \mathrm{C}=10^{-5} \mathrm{C} \\ & q_2=12 \mu \mathrm{C}=1.2 \times 10^{-5} \mathrm{C} \\ & r_1=10 \mathrm{~cm}=10^{-1} \mathrm{~m} \end{aligned}$

We have to find work done in bringing them

Closer when, $r_2=6 \mathrm{~cm}$

$=6 \times 10^{-2} \mathrm{~m}$

Work done, $W=$ change in potential energy in two configuration of charges

$\begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_2}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_1} \\ & =\frac{1}{4 \pi \varepsilon_0} q_1 q_2\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \\ & =9 \times 10^9 \times 10^{-5} \times 1.2 \times 10^{-5}\left(\frac{1}{6 \times 10^{-2}}-\frac{1}{10^{-1}}\right) \\ & =1.08\left(\frac{100}{6}-10\right)=\left(\frac{40}{6}\right) \times 1.08=7.2 \mathrm{~J} \end{aligned}$

Which is nearest to 8.1 J hence, option (a) is correct.

Given, charges at point A and B are $q_A,q_B=10\mu$C and $-10\mu$C

Separation between A and B = 10 cm = 10 $\times$ 10$^{-2}$ m

Let $O$ is mid-point of $A B$.

$\begin{aligned} \therefore \quad O A & =5 \times 10^{-2} \mathrm{~m} \Rightarrow O P=12 \times 10^{-2} \mathrm{~m} \\ \text { and } A P & =\sqrt{12^2+5^2} \\ & =\sqrt{169}=13 \times 10^{-2} \mathrm{~m} \end{aligned}$

As we know that, Electric field $(E)=\frac{k q}{r^2}$

where, $k$ is Coulomb's constant $=9 \times 10^9 \mathrm{C}^2 \mathrm{~m}^{-2} \mathrm{~N}^{-1}, q$ is charge and $r$ is separation between $A$ and $P$.

$E_A=\frac{9 \times 10^9 \times 10 \times 10^{-6}}{\left(13 \times 10^{-2}\right)^2}=5.3 \times 10^6 \mathrm{~N} / \mathrm{C}$

Similarly, electric field due to charge at $B, E_B=5.3 \times 10^6 \mathrm{~N} / \mathrm{C}$

$\therefore$ Net Electric field $\left(E_{\text {net }}\right)=2 E \sin \theta$

$\begin{aligned} & =2 \times 5.3 \times 10^6 \times \frac{5}{13}=\frac{53}{13} \times 10^6 \\ & =4.07 \times 10^6 \mathrm{~N} / \mathrm{C} \\ & =4.1 \times 10^6 \mathrm{~N} / \mathrm{C} \end{aligned}$

As we know that, charge is a conserved quantity i.e. charge of any isolated system remains same.

Hence, net charge of the system doesn’t change.

As we know that, on equipotential surface potential remains same everywhere and electric field lines are always perpendicular to equipotential surface.

Since, $E=-dV/dx$

= $-$ (Potential gradient)

$\therefore$ Angle between potential and potential gradient = $\frac{\pi}{2}$ rad

Given,

charges $q_1, q_2$ and $q_3$ are $-14 \mathrm{~nC}, 78.85 \mathrm{~nC}$ and $-56 \mathrm{~nC}$.

As we know that,

$\text { flux, } \phi=\frac{Q_{\text {enclosed }}}{\varepsilon_0}$

where, $\varepsilon_0$ is free space permittivity

$\begin{aligned} & =8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ \therefore \quad \phi_{\text {net }} & =\frac{(-14+78.85-56) \times 10^{-9}}{8.854 \times 10^{-12}} \\ & =10^3 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1} \end{aligned}$