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

An oil drop with a radius of $1 \, \mu \mathrm{m}$ is held stationary in a constant electric field of $3.65 \times 10^{4} \, \mathrm{N/C}$ due to excess electrons on it. The oil drop's density is $1.26 \, \mathrm{g/cm}^{3}$. To find the number of excess electrons on the oil drop, consider the following details:

Convert the given measurements:

Radius of the drop, $ r = 1 \, \mu \mathrm{m} = 10^{-6} \mathrm{~m} $.

Density of the oil, $\rho_{\text{oil}}$, is converted to:

$ \rho_{\text{oil}} = 1.26 \, \mathrm{g/cm}^{3} = 1.26 \times 10^{3} \, \mathrm{kg/m}^{3} $

Since the droplet is stationary, the gravitational force acting downward is balanced by the electric force acting upward:

$ \text{Weight of the droplet} = \text{Force due to the electric field} $

Mathematically, this can be expressed as:

$ \frac{4}{3} \pi r^{3} \rho_{\text{oil}} \cdot g = qE $

where $ q $ is the charge due to excess electrons. If $ n $ is the number of excess electrons, then:

$ q = ne $

Substitute this into the previous equation:

$ \frac{4}{3} \pi r^{3} \rho_{\text{oil}} \cdot g = neE $

Solving for $ n $:

$ n = \frac{4 \pi r^{3} \rho_{\text{oil}} g}{3eE} $

Substitute the known values into the equation to calculate $ n $:

$ \begin{aligned} n & = \frac{4 \times 3.14 \times (10^{-6})^{3} \times 1.26 \times 10^{3} \times 10}{3 \times 1.6 \times 10^{-19} \times 3.65 \times 10^{4}} \\ & = 9.03 \approx 9 \end{aligned} $

Thus, there are approximately 9 excess electrons on the oil drop.

When eight small liquid drops combine to form one large drop, the potential of the large drop is given as 20 V. We need to find the potential of each individual small drop.

Solution

Volume Relationship:

The total volume of the eight small drops is equal to the volume of the one large drop. This relationship is expressed as:

$ \left(\frac{4}{3} \pi r^3\right) \times 8 = \frac{4}{3} \pi R^3 $

Simplifying gives:

$ 2r = R \quad \ldots \text{(i)} $

This equation indicates the radius of the large drop $ R $ is twice the radius of one small drop $ r $.

Charge Conservation:

The total charge of the eight small drops is equal to the charge of the large drop:

$ 8q = Q \quad \ldots \text{(ii)} $

Potential Derivation:

The potential for one small drop ($ V' $) and the large drop ($ V $) are given by:

$ V' = \frac{q}{4 \pi \varepsilon_0 r} $

$ V = \frac{Q}{4 \pi \varepsilon_0 R} $

Finding the Ratio:

Using the above expressions, we derive:

$ \frac{V'}{V} = \frac{q}{Q} \times \frac{R}{r} $

Substituting from equations (i) and (ii), we get:

$ \frac{V'}{20} = \frac{q}{8q} \times \frac{2r}{r} $

This simplifies to:

$ \frac{V'}{20} = \frac{1}{4} $

Calculating $ V' $:

Solving for $ V' $, we find:

$ V' = 5 \, \text{V} $

Thus, the potential of each single small liquid drop is 5 V.

To determine how many electrons have been removed from a neutral dust particle suspended in an electric field, let's go through the calculations step by step.

Given:

Mass of the dust particle, $ m = 4 \times 10^{-12} \text{ mg} $

Conversion to kilograms:

$ m = 4 \times 10^{-12} \times 10^{-3} \text{ g} = 4 \times 10^{-12} \times 10^{-3} \times 10^{-3} \text{ kg} = 4 \times 10^{-18} \text{ kg} $

Electric field strength, $ E = 50 \text{ N/C} $

Gravitational acceleration, $ g = 10 \text{ m/s}^2 $

Calculate the Weight:

The weight $ W $ of the dust particle is given by:

$ W = m \times g = 4 \times 10^{-18} \times 10 = 4 \times 10^{-17} \text{ N} $

Equating Electric Force to Weight:

The electric force $ F_e $ experienced by the dust particle is:

$ F_e = q \times E $

Where $ q = n \times e $, with $ e = 1.6 \times 10^{-19} \text{ C} $ as the charge of one electron, and $ n $ being the number of electrons removed.

At equilibrium, the electric force equals the weight of the particle:

$ n \times e \times E = W $

$ n \times 1.6 \times 10^{-19} \times 50 = 4 \times 10^{-17} $

Solving for $ n $:

$ n = \frac{4 \times 10^{-17}}{1.6 \times 10^{-19} \times 50} $

$ n = \frac{400}{80} = 5 $

Therefore, 5 electrons were removed from the neutral dust particle.

The electric field $\mathbf{E}$ at point $(30, 30, 0)$ due to a charge of $0.008 \, \mu \mathrm{C}$ placed at the origin can be calculated using the formula:

$ \mathbf{E} = \frac{K q}{r^3} \cdot \mathbf{r} $

Step 1: Calculate the distance $r$:

$ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} $

Substitute the given coordinates:

$ r = \sqrt{30^2 + 30^2 + 0^2} = 30 \sqrt{2} \, \text{cm} = 30 \sqrt{2} \times 10^{-2} \, \text{m} $

Step 2: Convert charge $q$ to Coulombs:

$ q = 8 \times 10^{-3} \, \mu\mathrm{C} = 8 \times 10^{-3} \times 10^{-6} \, \text{C} $

Step 3: The position vector $\mathbf{r}$ is:

$ \mathbf{r} = (30 \hat{\mathbf{i}} + 30 \hat{\mathbf{j}}) \times 10^{-2} \, \text{m} $

Step 4: Substitute into the electric field formula:

$ \mathbf{E} = \frac{9 \times 10^9 \times 8 \times 10^{-3} \times 10^{-6}}{\left(30 \sqrt{2} \times 10^{-2}\right)^3} \times (30 \hat{\mathbf{i}} + 30 \hat{\mathbf{j}}) \times 10^{-2} $

Calculation:

$ \mathbf{E} = \frac{9 \times 8 \times 10^9 \times 10^{-11}}{27 \times 2 \sqrt{2} \times 10^{-6} \times 10^3} \times 30(\hat{\mathbf{i}} + \hat{\mathbf{j}}) $

Simplify the expression:

$ \mathbf{E} = \frac{9 \times 8 \times 10^2 \times 3}{27 \times 2 \sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) $

$ \mathbf{E} = 200 \sqrt{2}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \, \text{NC}^{-1} $

Thus, the electric field at the given point is $200 \sqrt{2}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \, \mathrm{NC}^{-1}$.

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 $\frac{R}{3}$.