Electrostatics

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 electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:

$ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} $

In this case, we know (V) and (Q), and we're asked to solve for (r). We can rearrange the formula to solve for (r):

$ r = \frac{1}{4\pi\epsilon_0} \frac{Q}{V} $

Substituting the given values into this equation gives:

$ r = \frac{9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}}{1} \frac{5 \times 10^{-9} \, \text{C}}{50 \, \text{V}} = 0.9 \, \text{m} $

So, the distance of the point P from the point charge is 0.9 meters. This corresponds to 90 cm.

${E_I} \ne 0$ (inside region)

${E_{II}} = 0$ (conducting region)

${E_{III}} \ne 0$

$ = {{KQ} \over {{r^2}}}\,\,\,(r > b)$

$\overrightarrow E = \left( {{A \over {{x^2}}}\widehat i + {B \over {{y^3}}}\widehat j} \right)$

$\left[ {{A \over {{x^2}}}} \right] = [E] = \left[ {{F \over q}} \right] = \left[ {{N \over C}} \right] = N{C^{ - 1}}$

$\mathrm{[A] = (N{m^2}{C^{ - 1}})}$

$\mathrm{[B] = N{m^3}{C^{ - 1}}}$

$\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi {{(2R)}^2}}}$

Now $Q{'_2} = {Q_{total}}\left[ {{{{R_2}} \over {{R_1} + {R_2}}}} \right]$

$ = ({Q_1} + {Q_2})\left[ {{{2R} \over {3R}}} \right]$

$ = \sigma (20\pi {R^2}){2 \over 3}$

$\therefore$ $\sigma {'_2} = {{Q{'_2}} \over {4\pi {{(2R)}^2}}} = {{\sigma (20\pi {R^2}){2 \over 3}} \over {16\pi {R^2}}}$

$ = {5 \over 4} \times {2 \over 3}\sigma $

$ = {5 \over 6}\sigma $

$w = \int {F\,.\,ds} $

$\overrightarrow F = q\overrightarrow E = 2 \times {10^{ - 2}} \times 30\widehat i = 0.6N\widehat i$

$w = \overrightarrow F .\overrightarrow d = (0.6\widehat i)\,.\,( - \widehat i, - 2\widehat j)$

$ = - 0.6$ J

$ = - 600$ mJ

Given,

Radius of spherical shell $=r$

Uniform charge density = 5

Size of cube edge $=3 r$

Let, spherical shell has charge $=q$

Change density $\sigma=\frac{q}{A}$

$ \begin{array}{rr} \sigma & =\frac{q}{4 \pi r^2} \\ q=\sigma 4 \pi r^2 & \ldots \text { (i) }[\mathrm{C} \end{array} $

[Charge on shell]

flux through one face of cube;

$ \phi_E=\frac{1}{6} \frac{q}{\varepsilon_0} $

Put the value of $q=\frac{1}{6} \times \frac{4 \pi r^2 \sigma}{\varepsilon_0}$

$ \phi_E=\frac{2 \pi r^2 \sigma}{3 \varepsilon_0} $

Hence, the correct option is (d).

Given, for a hollow conducting sphere,

Radius $=R$

Electric potential at the surface, $V_{r=R}=V$

∴ Electric potential at $r=\frac{R}{3}, V_{r=R / 3}=$ ?

In, hollow conducting sphere, the charge will accumulate on the surface.

Electric field inside the sphere is zero.

Using relation,

$ \begin{aligned} E & =\frac{-\partial V}{\partial x} \\ \Rightarrow \quad O & =-\partial V \end{aligned} $

$\therefore V$ is constant inside the sphere and is equal to value of potential on the surface of sphere.

Hence, potential will be $V$ for $\frac{R}{3}$.

Let two electrons be repelled by electrostatic force and attracted by their gravitational forces.

Therefore, $F_G=\frac{G m_e m_e}{r^2}$

and $ F_{\mathrm{EM}}=\frac{K e}{r^2} $

$ \begin{aligned} \frac{F_6}{F_{\mathrm{EM}}} & =\frac{G \times 9.11 \times 10^{-31} \times 9.11 \times 10^{-31}}{K \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}} \\ & =\frac{6.67 \times 10^{-11} \times 9.11 \times 9.11 \times 10^{-62}}{9 \times 10^9 \times 1.6 \times 1.6 \times 10^{-38}} \\ & \approx 10^{-37} \end{aligned} $

$ \begin{aligned} &\text { According to given situation, }\\ &\begin{aligned} & & \sigma_1 & =\sigma_2 \\ & & \frac{Q_1}{4 \pi r_1^2} & =\frac{Q_2}{4 \pi r_2^2} \\ & \Rightarrow & \frac{Q_1}{4 \pi \varepsilon_0 r_1^2} & =\frac{Q_2}{4 \pi \varepsilon_0 r_2^2} \\ & \Rightarrow & E_1 & =E_2 \\ \therefore & & E_1: E_2 & =1: 1 \end{aligned} \end{aligned} $

The given situation is shown below.

Potential at point $P$ is zero.

i.e.

$ \begin{aligned} & \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 \times 10^{-6}}{x}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\left(-4 \times 10^{-6}\right)}{3-x}=0 \\ & \Rightarrow \quad \frac{1}{x}-\frac{3}{3-x}=0 \\ & \Rightarrow \quad \frac{1}{x}=\frac{2}{3-x} \\ & \Rightarrow \quad 3-x=2 x \Rightarrow x=1 \mathrm{~m} \end{aligned} $

Electric field at point $P$,

$ \begin{aligned} E_p=\frac{1}{4 \pi \varepsilon_0} & \frac{2 \times 10^{-6}}{(x)^2}-\frac{4 \times 10^{-6}}{4 \pi \varepsilon_0 \cdot(3-x)^2} \\ & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{2 \times 10^{-6}}{(1)^2}-\frac{4 \times 10^{-6}}{(2)^2}\right] \\ & =9 \times 10^9 \times 10^{-6} \\ & =9000 \mathrm{NC}^{-1} \end{aligned} $

Given, $\mathbf{E}=24 \hat{\mathbf{i}}+30 \hat{\mathbf{j}}+28 \hat{\mathbf{k}} \mathrm{~N} / \mathrm{C}$

$ A=20 \mathrm{~m}^2 $

Since, area lies in $Y Z$-plane

Hence, $\mathbf{A}=A \hat{\mathbf{i}}=20 \hat{\mathbf{i}}$

∴ Electric flux, $\phi=\mathbf{E} \cdot \mathbf{A}$

$ \begin{aligned} & =(24 \hat{\mathbf{i}}+30 \hat{\mathbf{j}}+28 \hat{\mathbf{k}}) \cdot(20 \hat{\mathbf{i}}) \\ & =480 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1} \end{aligned} $