Electrostatics

Effective air distance between charges is,

$ d=(r-t)+t \sqrt{K} $

$r=$ distance between charges

$t=$ thickness of dielectric slab

$K=$ dielectric constant

∴ Force is $F=\frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{(r-t+t \sqrt{K})^2}$

$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q_1 Q_2}{r^2}\left(\frac{r}{r-t+t \sqrt{K}}\right)^2 \\ & =F_{\text {air }}\left(\frac{r}{r-t+t \sqrt{K}}\right)^2 \end{aligned} $

Here, $t=30 \%$ of $r=\frac{30}{100} r$

and $F=\frac{F_{\text {air }}}{2.56}$

$ \begin{aligned} & \text { So, } \frac{1}{2.56}=\left(\frac{r}{r-\frac{30}{100} r+\frac{30}{100} r \times \sqrt{K}}\right)^2 \\ & \Rightarrow \quad \frac{10}{16}=\frac{r}{r-0.3 r+0.3 r \sqrt{K}} \\ & \Rightarrow \quad \frac{10}{16}=\frac{1}{0.7+0.3 \sqrt{K}} \\ & \Rightarrow \quad 7+3 \sqrt{K}=16 \\ & \quad 3 \sqrt{K}=9 \\ & \quad \sqrt{K}=3 \text { or } K=9 . \end{aligned} $

$729 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3$

$ \begin{aligned} \Rightarrow \quad(9 r)^3 & =R^3 \\ R^3 & =9 r \end{aligned} $

Potential on small sphere,

$ \begin{aligned} & \boldsymbol{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ \Rightarrow & 3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ \Rightarrow & \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Charge on big sphere, $Q=729 q$

∴ Potential on big sphere,

$ V^{\prime}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R} $

$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{729 q}{9 r}=81 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \\ & =81 \times 3 \\ & =243 \mathrm{volt} \\ & \text { [From Eq. (i)] } \end{aligned} $

Step 1: Formula for electric field around a long wire
The electric field ($E$) at a distance $x$ from a very long straight wire with charge per unit length ($\lambda$) is given by:

$E = \frac{2 K \lambda}{x}$
Here, $K$ is Coulomb's constant ($K = 9 \times 10^9~\mathrm{N\,m^2\,C^{-2}}$).

Step 2: Substitute the given values
We are given:

• $E = 7.5 \times 10^4~\mathrm{N/C}$
• $\lambda = 2.5 \times 10^{-7}~\mathrm{C/m}$
• $K = 9 \times 10^9~\mathrm{N\,m^2\,C^{-2}}$

Step 3: Solve for $x$
Multiply the numbers on the right side first:

• $2 \times 9 = 18$
• $18 \times 2.5 = 45$
• So, top becomes: $45 \times 10^9 \times 10^{-7} = 45 \times 10^{2} = 4500$

Step 4: Rearrange to find $x$
Multiply both sides by $x$ and divide both sides by $7.5 \times 10^4$: $x = \frac{4500}{7.5 \times 10^4} = \frac{4500}{75000} = 0.06~\mathrm{m}$

Step 5: Convert meters to centimeters
$0.06~\mathrm{m} = 6~\mathrm{cm}$

Step 1: Displacement equations

The proton and the alpha particle both start from rest, so their displacement after some time is given by:

$s = \frac{1}{2} a t^2.$

Since they travel the same distance:

$\frac{1}{2} a_\alpha t_\alpha^2 = \frac{1}{2} a_p t_p^2$

This means:

$a_\alpha t_\alpha^2 = a_p t_p^2$

Step 2: Acceleration in an electric field

The acceleration of a charged particle in an electric field $E$ is:

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

- For a proton, charge $= e$, mass $= m_p$

- For an alpha particle, charge $= 2e$, mass $= 4m_p$

So: - $a_p = \frac{eE}{m_p}$ - $a_\alpha = \frac{2eE}{4m_p} = \frac{eE}{2m_p}$

Step 3: Substitute accelerations into equation

Plug the accelerations into the earlier equation:

$\left(\frac{e E}{2 m_p}\right) t_\alpha^2 = \left(\frac{e E}{m_p}\right) t_p^2$

Step 4: Solve for time ratio

Divide both sides by $\frac{eE}{m_p}$:

$\frac{t_\alpha^2}{2} = t_p^2$

This gives:

$t_\alpha^2 = 2 t_p^2$

Taking square roots:

$t_\alpha = \sqrt{2} t_p$

So,

$\frac{t_p}{t_\alpha} = \frac{1}{\sqrt{2}}$

or

$t_p : t_\alpha = 1 : \sqrt{2}$

$ \begin{aligned} &\text { Net force on the charge at origin }=0\\ &\begin{aligned} & \Rightarrow \frac{1}{4 \pi \varepsilon_0}\left[\begin{array}{l} \frac{2 \times 10^{-6} Q}{\left(1 \times 10^{-2}\right)^2}+\frac{2 \times 10^{-6} \times 4 \times 10^{-6}}{\left(2 \times 10^{-2}\right)^2} \\ +\frac{2 \times 10^{-6} \times 12 \times 10^{-6}}{\left(4 \times 10^{-2}\right)^2} \end{array}\right]=0 \\ & \Rightarrow \frac{Q}{10^{-4}}+\frac{4 \times 10^{-6}}{4 \times 10^{-4}}+\frac{12 \times 10^{-6}}{16 \times 10^{-4}}=0 \\ & \Rightarrow Q+10^{-6}+\frac{3}{4} \times 10^{-6}=0 \\ & \Rightarrow Q+\frac{7}{4} \times 10^{-6}=0 \\ & \Rightarrow Q=-\frac{7}{4} \times 10^{-6} \mathrm{C}=-\frac{7}{4} \mu \mathrm{~F}=-1.75 \mu \mathrm{~F} \end{aligned} \end{aligned} $

According to work energy theorem,

$ \begin{aligned} & W=q \Delta V \\ & \mathrm{KE}_f-\mathrm{KE}_i=q \Delta V \\ & \frac{1}{2} m v^2-0=2 \times 10^{-6} \times 160 \\ & \Rightarrow \frac{1}{2} \times 10^{-5} v^2=3.2 \times 10^{-4} \\ & \Rightarrow v^2=64 \Rightarrow v=8 \mathrm{~m} / \mathrm{s} \end{aligned} $

Acceleration of electron in electric field is

$ a_e=\frac{F_e}{m}=-\frac{e E}{m} $

Acceleration of positron in electric field.

$ a_p=\frac{F_p}{m}=\frac{e E}{m} $

Displacement of electron in electric field

$ y_e=0 \times t+\frac{1}{2} a_e t^2=-\frac{1}{2} \frac{e E}{m} t^2 $

Displacement of positron in electric field

$ y_p=\frac{1}{2} a_p t^2=\frac{1}{2} \frac{e E}{m} t^2 $

The distance of separation between electron and positron

$ \begin{aligned} & =\left|y_p-y_e\right| \\ & =\left|\frac{\frac{1}{2} e E t^2}{m}-\left(-\frac{1}{2} \frac{e E t^2}{m}\right)\right| \\ & =\left|\frac{e E t^2}{m}\right|=\frac{e E t^2}{m} \end{aligned} $

The potential at points $B$ and $C$ due to the charges at $O$ and $A$

Potential at $B$

$ V_B=\frac{q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 2 R} $

Potential at $C$

$ V_c=\frac{q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 \sqrt{2} R} $

Work done

$ \begin{aligned} W & =q\left(V_C-V_B\right) \\ & =q\left[\begin{array}{l} \left(\frac{q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 \sqrt{2} R}\right) \\ -\left(\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{4 \pi \varepsilon_0 2 R}\right) \end{array}\right] \\ & =\frac{q^2}{4 \pi \varepsilon_0 R}\left[\frac{1}{\sqrt{2}}-\frac{1}{2}\right] \\ & =\frac{q^2}{4 \pi \varepsilon_0 R}\left[\frac{\sqrt{2}-1}{2}\right] \end{aligned} $

The electromagnetic force between two protons is much stronger than the gravitational force between them.

Let $F_e$ be the electromagnetic force, and $F_g$ be the gravitational force.

According to the problem: $ F_e = 10^n F_g $

We write the formulas for both forces:

Electromagnetic force: $ F_e = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} $

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

Replace $F_e$ and $F_g$ in the equation: $ \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = 10^n \frac{G m^2}{r^2} $

Since $r^2$ is on both sides, it cancels out:

$ \frac{1}{4\pi\varepsilon_0} e^2 = 10^n G m^2 $

Solve for $10^n$:

$ 10^n = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{G m^2} $

Now, use the constants:

$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$

Charge of proton, $e = 1.6 \times 10^{-19}$

Gravitational constant, $G = 6.67 \times 10^{-11}$

Mass of proton, $m = 1.67 \times 10^{-27}$

Plug in the values:

$ 10^n = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2} $

Calculate the numerator: $9 \times 10^9 \times (1.6 \times 10^{-19})^2 = 9 \times 10^9 \times 2.56 \times 10^{-38} = 2.304 \times 10^{-28}$

Calculate the denominator: $6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2 = 6.67 \times 10^{-11} \times 2.79 \times 10^{-54} = 1.86 \times 10^{-64}$

Divide the values: $ \frac{2.304 \times 10^{-28}}{1.86 \times 10^{-64}} = 1.23 \times 10^{36} $

This means $10^n = 1.23 \times 10^{36}$, so $n = 36$.

Total charge on spherical shell.

$ Q=\sigma A=\sigma\left(4 \pi R^2\right)=4 \pi R^2 \sigma $

According to Gauss's law total electric flux through a closed surface.

$ \phi=\frac{Q}{\varepsilon_0}=\frac{4 \pi R^2 \sigma}{\varepsilon_0} $

$\therefore$ Flux through one face

$ \phi_1=\frac{\phi}{6}=\frac{4 \pi R^2 \sigma}{6 \varepsilon_0}=\frac{2 \pi R^2 \sigma}{3 \varepsilon_0} $

Given,

$ \begin{array}{l} \mathbf{E}=(8 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C} \\\\ A=3 \mathrm{~m}^{2} \\\\ \phi=? \end{array} $

$ \therefore $ Area is lying in the $ X Z $-plane, so vector component of area,

$ \mathbf{A}=3 \hat{\mathbf{j}} $

Now, $\phi=\mathbf{E} \cdot \mathbf{A}$

$ \begin{aligned} & \phi=(8 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}) \cdot(3 \hat{\mathbf{j}}) \\\\ & \phi=(0+39) \mathrm{Wb} \\\\ & \phi=39 \mathrm{~Wb} \end{aligned} $

The work done by an electric field on a charged particle is determined by the equation $ W = F \cdot s $, where $ F $ is the force exerted by the electric field and $ s $ is the displacement of the particle.

The force $ F $ can be expressed as $ F = qE $, where $ q $ is the charge of the particle and $ E $ is the electric field strength. The displacement $ s $ can be determined using the kinematic equation $ s = \frac{1}{2} a t^2 $, where $ a $ is the acceleration and $ t $ is the time.

Substituting these into the work formula gives:

$ W = E \cdot q \cdot \frac{1}{2} a t^2 $

Since both the proton and the alpha particle are in the same electric field and experienced for the same time:

$ W \propto q \cdot a $

Let's denote:

$ W_1 $ as the work on the proton,

$ W_2 $ as the work on the alpha particle.

Hence,

$ \frac{W_1}{W_2} = \frac{q_1 \cdot a_1}{q_2 \cdot a_2} $

For the proton:

$ q_1 = 1 \, \text{C} $,

The acceleration $ a_1 = \frac{E \cdot q_1}{m} $ (where $ m $ is the mass of the proton).

For the alpha particle:

$ q_2 = 2 \, \text{C} $,

The acceleration $ a_2 = \frac{E \cdot (2q)}{4m} = \frac{E}{2} $.

Substituting these into the work ratio equation yields:

$ \frac{W_1}{W_2} = \frac{1 \cdot \frac{E}{E}}{2 \cdot \frac{E}{2E}} $

Simplifying,

$ \frac{W_1}{W_2} = \frac{1}{2} \times 2 = 1:1 $

Thus, the ratio of the work done by the electric field on the proton compared to the alpha particle is $ 1:1 $.

$ Q_{1}=-10 \mu \mathrm{C}, Q_{2}=+5 \mu \mathrm{C} $

$x=0, x=\sqrt{2} \mathrm{~m}$

Net electric field is zero at point $ P $

$ \begin{array}{c} E=\frac{k\left[Q_{1}\right]}{x^{2}}+\frac{k\left[Q_{2}\right]}{(x-\sqrt{2})^{2}}=0 \\\\ \frac{(-10 \mu \mathrm{C}) k}{x^{2}}+\frac{k(5 \mu \mathrm{C}}{(x-\sqrt{2})^{2}}=0 \\\\ \frac{+2}{x^{2}}=\frac{1}{(x-\sqrt{2})^{2}} \end{array} $

Square root both side

$ \begin{aligned} \frac{\sqrt{2}}{x} & =\frac{1}{(x-\sqrt{2})} \\\\ \sqrt{2}(x-\sqrt{2}) & =x \\\\ x(\sqrt{2}-1) & =2 \\\\ x & =\frac{2}{\sqrt{2}-1} \end{aligned} $

Rationalize the denominator

$ \begin{array}{l} x=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\\\ x=2(\sqrt{2}+1) \mathrm{m} \end{array} $

From the diagram,

$ \frac{K(8)}{x^2}=\frac{K(32)}{(15+x)^2} \text { (direction of } $

attraction/repulsion is considered)

$ \begin{aligned} & \Rightarrow \quad \frac{1}{x^2}=\frac{4}{(15+x)^2} \\\\ & \Rightarrow \quad \frac{1}{x}=\frac{2}{15+x} \\\\ & \Rightarrow \quad 15+x=2 x \\\\ & \Rightarrow \quad x=15 \mathrm{~cm} \end{aligned} $

From $-8 \mu \mathrm{C}$ charge

Given, Distance of separation, $r=4 \mathrm{~m}$

$ r^2=16 \mathrm{~m} $

Sum of two charges

$ \begin{aligned} q_1+q_2 & =36 \mu \mathrm{C} \\\\ & =36 \times 10^{-6} \mathrm{C} \\\\ F & =0.18 \mathrm{~N} \end{aligned} $

Coulomb's law,

$ F=K \frac{q_1 q_2}{r^2} $

Here, $q_2=36 \times 10^{-6}-q_1$

Thus, $0.18=8.98 \times 10^9 \times \frac{q_1\left(36 \times 10^{-6}-q_1\right)}{16}$

Solving quadratic equation,

$ q_1=\frac{36 \times 10^{-6}+}{\sqrt{\left(36 \times 10^{-6}\right)^2-4 \times \frac{0.18 \times 16}{8.98 \times 10^9}}} $

$\begin{aligned} q_1= & \frac{36 \times 10^{-6}+}{\sqrt{\left(36 \times 10^{-6}\right)^2-4 \times \frac{2.88}{8.98 \times 10^9}}} \\\\ q_1 & =1.98 \times 10^{-5} \mathrm{C} \\\\ q_1 & \approx 20.8 \times 10^{-6} \mathrm{C} \\\\ q_1 & =20 \mu \mathrm{C}\end{aligned}$

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} $

To determine the magnitude of the charge, we have the following given information:

Electric field $E = 500 \, \mathrm{N/C}$

Electric potential $V = 30 \, \mathrm{V}$

First, we use the formula for the potential gradient, which relates the electric field and potential through the distance $x$ from the point charge:

$ E = \frac{V}{x} $

Solving for $x$:

$ x = \frac{30}{500} = 0.06 \, \mathrm{m} $

Now, we employ the formula for the electric potential due to a point charge:

$ V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{x} $

Substitute the known values to solve for the magnitude of the charge $q$:

$ 30 = \frac{9 \times 10^9}{0.06} \times q $

Rearranging gives:

$ q = \frac{30 \times 0.06}{9 \times 10^9} = 2 \times 10^{-10} \, \mathrm{C} $

Thus, the magnitude of the charge is:

$ 2 \times 10^{-10} \, \mathrm{C} $

Given,

' 'The charges corresponding to number's of a clock $-q,-2 q \ldots 12 q$.

Electric field is given by

$ E=k q / r^2 \Rightarrow E \propto q $

Let's take a pair of charges ( $-9 q$ and $-3 q$ )

The value of $E$ due to $-9 q>-3 q$

and net magnitude $=-6 q=E_0$

Same goes for all pairs with magnitude of $(-6 q)$.

The angle between each field,

$ \theta=\frac{360^{\circ}}{12}=30^{\circ} $

From the property of vector addition, if two equal vector is separated by $120^{\circ}$, then resultant vector is same as initial vector.

For making $120^{\circ}$, take vector 1 to vector 5 Resultant of vector 1 and $5=E_0=$ Vector 3 direction.

Resultant of vector 2 and $6=E_0=$ Vector 4 direction

Now, vector (1 and 5) and (2 and 6) are eliminated

Now, vector $3=E_0+E_0=2 E_0$

Vector $4=E_0+E_0=2 E_0$

Vector 3 denotes 10 AM

Vector 4 denotes 9 AM

So, the resultant vector of 9 AM and 10 AM is in midway direction.

i.e., $9: 30$.

The electric field $\vec{E}$ is related to the electric potential $V$ by the following relation:

$\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$

Given the potential $V = \frac{1}{2}(y^2 - 4x)$, we need to find the partial derivatives with respect to $x$ and $y$.

Partial derivative with respect to $ x $:

$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x} \left[\frac{1}{2}(y^2 - 4x)\right] = \frac{1}{2}(-4) = -2$

Partial derivative with respect to $ y $:

$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} \left[\frac{1}{2}(y^2 - 4x)\right] = \frac{1}{2}(2y) = y$

Now, let's plug these into the formula for the electric field:

$\vec{E} = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}\right) = -(-2\hat{i} + y\hat{j}) = 2\hat{i} - y\hat{j}$

We need to find the electric field at the point $x = 1\,\text{m}$ and $y = 1\,\text{m}$. So, we substitute $y = 1$ into the equation:

$\vec{E} = 2\hat{i} - (1)\hat{j} = (2\hat{i} - \hat{j})\,\text{V/m}$

So, the correct answer is:

Option C: $(2\hat{i} - \hat{j})\,\text{V/m}$

The given situation is shown in the figure below.

Since, block remains at rest on inclined plane, then

$ \begin{array}{rlrl} & q E =m g \sin 60^{\circ} \\ \Rightarrow & E =\frac{m g \sin 60^{\circ}}{q} \end{array} $

Here, $q=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}$

$ \begin{aligned} & m=5 \mathrm{~g}=5 \times 10^{-3} \mathrm{~kg} \\ \Rightarrow \quad g & =10 \mathrm{~m} / \mathrm{s}^2 \\ \therefore \quad E & =\frac{5 \times 10^{-3} \times 10}{5 \times 10^{-6}} \times \frac{\sqrt{3}}{2} \\ & =\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C} \end{aligned} $

Given, ratio of radii of two metal sphere,

$ \begin{aligned} & & R_1: R_2=4: 7 \\ \Rightarrow & & \frac{R_1}{R_2}=\frac{4}{7} \end{aligned} $

If $q_1$ and $q_2$ be the charges on both the spheres after being contact with charge of $8.8 \times 10^{-7} \mathrm{C}$. then $q_1+q_2=8.8 \times 10^{-7} \mathrm{C}$

$ \begin{aligned} & =0.88 \mu \mathrm{C} \\ & \Rightarrow \quad q_1+q_2=0.88 \mu \mathrm{C} \end{aligned} $

If $V$ be the common potential on both the spheres after connection, then

$ \begin{array}{lll} & & q_1=C_1 V \\ \text { and } & & q_2=C_2 V \\ \therefore & & \frac{q_1}{q_2}=\frac{C_1 V}{C_2 V}=\frac{C_1}{C_2}=\frac{4 \pi \varepsilon_0 R_1}{4 \pi \varepsilon_0 R_2} \\ \Rightarrow & & \frac{q_1}{q_2}=\frac{R_1}{R_2}=\frac{4}{7} \\ \Rightarrow & & q_1=\frac{4}{7} q_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii), we get

$ \begin{array}{ll} & \frac{4}{7} q_2+q_2=0.88 \\ \Rightarrow \quad & q_2=\frac{7}{11} \times 0.88=0.56 \mu \mathrm{C} \\ \therefore \quad & q_1=0.88-q_1 \\ & =0.88-0.56=0.32 \mu \mathrm{C} \end{array} $

i.e charge on smaller sphere,

$ q_1=0.32 \times 10^{-6} \mathrm{C} $

∴ Potential due to smaller sphere at $r=60 \mathrm{~m}$

$ \begin{aligned} V & =9 \times 10^9 \frac{q_1}{r} \\ & =9 \times 10^9 \times \frac{0.32 \times 10^{-6}}{60}=48 \mathrm{~V} \end{aligned} $

For the given situations, arrangement of charges are given as

The magnitude of force on charge, $q_3=5 \mu \mathrm{C}$

placed at mid-point $O$ is given as

$ \begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{(A O)^2}+\frac{1}{4 \pi \varepsilon_0} \frac{q_3 q_2}{(O B)^2} \\ & =9 \times 10^9 \times \frac{10 \times 10^{-6} \times 5 \times 10^{-6}}{\left(5 \times 10^{-2}\right)^2} \\ & -9 \times 10^9 \times \frac{5 \times 10^{-6} \times 10 \times 10^{-6}}{\left(5 \times 10^{-2}\right)^2} \end{aligned} $

$ \begin{aligned} & =\frac{45 \times 10^{-2}}{25 \times 10^{-4}}+\frac{45 \times 10^{-2}}{25 \times 10^{-4}} \\ & =\frac{9}{5} \times(10)^2+\frac{9}{5} \times(10)^2 \\ & =\frac{18}{5} \times(10)^2 \\ & =3.6 \times 10^2 \mathrm{~N} \\ & =360 \mathrm{~N} \end{aligned} $

Capacitance of sphere-1

$ C_1=4 \pi \varepsilon_0 R $

Capacitance of sphere-2

$ \begin{aligned} & C_2=4 \pi \varepsilon_0(3 R) \\ & C_2=12 \pi \varepsilon_0 R \end{aligned} $

We know after connection, charge flows from sphere-1 to sphere-2 till both acquired same potential. If $q_1{ }^{\prime}$ and $q_2{ }^{\prime}$ be charges on sphere-1 and sphere-2 respectively after connection, then

$ \frac{q_1^{\prime}}{q_2^{\prime}}=\frac{C_1 V}{C_2 V} $

(where, $V=$ common potential)

$ \Rightarrow \quad \frac{q_1^{\prime}}{q_2^{\prime}}=\frac{C_1}{C_2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The ratio of charge densities of sphere-1 and sphere-2 after connection is given as

$ \frac{\sigma_1}{\sigma_2}=\frac{\frac{q_1^{\prime}}{4 \pi R^2}}{\frac{q_2^{\prime}}{4 \pi(3 R)^2}}=\frac{q_1^{\prime}}{q_2^{\prime}} \times 9=\frac{C_1}{C_2} \times 9 $

[from Eq. (i)]

$ \therefore \quad \frac{\sigma_1}{\sigma_2}=9 \frac{C_1}{C_2}=9 \cdot \frac{4 \pi \varepsilon_0 R}{4 \pi \varepsilon_0(3 R)}=3 $

Charge placed on the centre of cube, $q=6 \mu \mathrm{C}=6 \times 10^{-6} \mathrm{C}$

According to Gauss's law, total electric flux passing through cube, $\phi=q / \varepsilon_0$ Since, there are 6 faces in cube. Hence, electric flux passing through each face,

$ \begin{aligned} \phi_1 & =\frac{\phi}{6} \\ & =\frac{q}{6 \varepsilon_0}=\frac{q}{\frac{3}{2 \pi} \times 4 \pi \varepsilon_0} \\ & =\frac{2 \pi}{3} \times q \cdot \frac{1}{4 \pi \varepsilon_0} \\ & =\frac{2 \pi q}{3} \times 9 \times 10^9 \\ & =\frac{2 \pi}{3} \times 6 \times 10^{-6} \times 9 \times 10^9 \\ & =36 \pi \times 10^3 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C} \end{aligned} $

The given situation is shown below.

Here, $x=\sqrt{3} R$

Now, total potential at the centre of the Ring-1.

$V_1=$ potential due to Ring-1 + Potential due to Ring-2.

$ =\frac{q}{4 \pi \varepsilon_0 R}+\frac{(-q)}{4 \pi \varepsilon_0 \sqrt{\left(R^2+x^2\right)}} $

Similarly, total potential at the centre of the Ring-2.

$ V_2=\frac{-q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 \sqrt{R^2+x^2}} $

Hence, Potential difference,

$ \begin{aligned} & V_1-V_2=2\left(\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{4 \pi \varepsilon_0 \sqrt{R^2+x^2}}\right) \\ &=2\left(\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{4 \pi \varepsilon_0 \sqrt{R^2+(\sqrt{3} R)^2}}\right) \\ & {[\because x=\sqrt{3} R] } \\ &=2\left[\frac{q}{4 \pi \varepsilon_0 R}-\frac{q}{8 \pi \varepsilon_0 R}\right] \\ &=2\left[\frac{q}{8 \pi \varepsilon_0 R}\right]=\frac{q}{4 \pi \varepsilon_0 R} \end{aligned} $

Given, mass of charges,

$ \begin{aligned} m_1 & =m_2=1 \mathrm{~g}=10^{-3} \mathrm{~kg} \\ r & =1 \mathrm{~m} \end{aligned} $

∴ Gravitational force between them ,

$ \begin{aligned} F_G & =\frac{G m_1 m_2}{r^2} \\ & =\frac{6.67 \times 10^{-11} \times 10^{-3} \times 10^{-3}}{1^2} \\ & {\left[\therefore G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\right] } \\ & =6.67 \times 10^{-17} \mathrm{~N} \end{aligned} $

Again, charge on $m_1$ and $m_2$,

$ q_1=q_2=1 \text { femto coulomb }=10^{-15} \mathrm{C} $

∴ Electrostatic force between the both charges,

$ \begin{aligned} F_e & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \\ & =9 \times 10^9 \times \frac{10^{-15} \times 10^{-15}}{1^2} \\ & =9 \times 10^{-21} \mathrm{~N} \end{aligned} $

∴ Clearly $F_G \gg F_e$

Hence, gravitational force is dominant force.

The given arrangement of system of charges are shown as

$ \text { In } \triangle A B C, \sin 30^{\circ}=\frac{B C}{A C} $

$ \Rightarrow \quad \frac{1}{2}=\frac{B C}{10} \Rightarrow B C=5 \mathrm{~cm} $

Similarly, $A B=A C \sin 60^{\circ}=10 \times \frac{\sqrt{3}}{2}$

$ =5 \sqrt{3} \mathrm{~cm} $

The given charges form two electric dipole of dipole moment $p_1$ and $p_2$ as shown below

Since, $p_1=A B \times q=5 \sqrt{3} q(\mathrm{C}-\mathrm{m})$

and $p_2=B C \times q=5 q \mathrm{C}-\mathrm{m}$

The magnitude of dipole moment of the combination

$ \begin{aligned} p & =\sqrt{p_1^2+p_2^2} \\ & =\sqrt{(5 \sqrt{3} q)^2+(5 q)^2} \\ \Rightarrow \quad p & =\sqrt{100 q^2}=10 q \end{aligned} $

The given situation is shown below

Acceleration of the charge particle in electric field,

$ a_y=\frac{F}{m}=\frac{q E}{m} $

Time taken by the charged particle to cross the field, $t=x / v$.

Upward component of velocity of charge particle emerging, from field region,

$ \begin{aligned} v_y & =u_y t+\frac{1}{2} a_y t^2 \\ & =0 \times t+\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{x}{v}\right)^2 \\ v_y & =\frac{1}{2} \frac{q E x^2}{m v^2} \end{aligned} $

This equation is in the form of $y=k x^2$, which represents the equation of parabola.

Given, $\sigma=8.85 \times 10^{-6} \mathrm{C} / \mathrm{m}^2$

Kinetic energy of electron,

$ K=8 \times 10^{-17} \mathrm{~J} $

We know that, electric field due to metal plate is given by

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

∴ Force on electron in this electric field,

$ F=-e E=-\frac{e \sigma}{\varepsilon_0} $

∴ Acceleration on electron,

$ a=\frac{F}{m}=-\frac{e \sigma}{m \varepsilon_0} $

Let $u$ be the initial velocity of electron, then

$ v^2=u^2+2 a s $

but $v=0$

$ \begin{aligned} &\begin{aligned} & \therefore \quad 0=u^2+2\left(\frac{-e \sigma}{m \varepsilon_0}\right) s \\ & \Rightarrow \quad s=\frac{u^2 \varepsilon_0 m}{2 e \sigma} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { but } \quad K=\frac{1}{2} m u^2 \\ & \Rightarrow \quad u^2=\frac{2 K}{m} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned}\\ &\text { ∴ From Eqs. (i) and (ii), we get }\\ &s=\frac{2 K \varepsilon_0 m}{2 m e \sigma}=\frac{K \varepsilon_0}{e \sigma} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad s & =\frac{8 \times 10^{-17} \times 8.85 \times 10^{-12}}{1.6 \times 10^{-19} \times 8.85 \times 10^{-6}} \\ & =5 \times 10^{-4} \mathrm{~m}=0.5 \mathrm{~mm} \end{aligned} $

The given situation is shown below.

When electron reaches at point $B$, then change in its potential energy is equal to gain in kinetic energy.

i.e. $\Delta U=K$

$\Rightarrow \mathrm{PE}$ at point $B-\mathrm{PE}$ at point $A=\frac{1}{2} m_e v^2$

$ \begin{aligned} & \Rightarrow \quad \frac{1}{4 \pi \varepsilon_0} \frac{q_1 \cdot q}{r_2^2}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q}{r_1^2}=\frac{1}{2} m_e v^2 \\ & \Rightarrow \quad \frac{1}{4 \pi \varepsilon_0}\left[\frac{e q}{r_2^2}-\frac{e q}{r_1^2}\right]=\frac{1}{2} m_e v^2 \\ & \Rightarrow \quad 9 \times 10^9\left[\frac{e q}{2^2}-\frac{e q}{4^2}\right]=\frac{1}{2} m_e v^2 \\ & \Rightarrow \quad 9 \times 10^9\left(\frac{3 e q}{16}\right)=\frac{1}{2} m_e v^2 \\ & \Rightarrow v=\sqrt{\frac{3 e q \times 9 \times 10^9}{8 m_e}} \\ & \Rightarrow v=\sqrt{\frac{3 \times 1.6 \times 10^{-19} \times 2 \times 10^{-8} \times 9 \times 10^9}{8 \times 9 \times 10^{-31}}} \\ & =3.46 \times 10^6 \end{aligned} $

Which is nearest to $4 \times 10^6 \mathrm{~m} / \mathrm{s}$.

According to Gauss's law, the electric flux through a closed surface is Electric flux, $\phi=\frac{Q}{\varepsilon_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Electric flux through a closed surface does not depend on the dimensions of closed surface, it depends only the amount of charge enclosed by the closed surface.

According to question,

$ \begin{aligned} Q^{\prime}=2 Q \Rightarrow \phi^{\prime} & =\frac{Q^{\prime}}{\varepsilon_0}=\frac{2 Q}{\varepsilon_0} \\ & =2 \phi \text { [using Eq. (i)] } \end{aligned} $

Given, $K=\frac{4}{3}$

For a dielectric, the electric displacement,

$ \begin{aligned} \mathbf{D} & =\varepsilon_0 \mathbf{E}+\mathbf{P} \text { or } D=\varepsilon_0 E+P \\ K \varepsilon_0 E & =\varepsilon_0 E+\chi E \Rightarrow \chi=(K-1) \varepsilon_0 \\ \Rightarrow \quad \chi & =\varepsilon_0\left(\frac{4}{3}-1\right) \Rightarrow \chi=\frac{\varepsilon_0}{3} \end{aligned} $

Electric field due to a uniformly charged non-conducting solid sphere of radius $R$ is given as

$ E= \begin{cases}\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q r}{R^3} \quad ; \quad \text { for } r

Hence, graph for electric field versus distance $r$ for a non-conducting solid sphere is given as

The given situation is shown in the following figure.

Work done in moving a charge $Q$ along semi-circle is equal to change in potential energy.

i.e.

$ \begin{aligned} W= & U_{\text {final }}-U_{\text {initial }} \\ = & {\left[\frac{k q Q}{A D}+\frac{k(-q) Q}{B D}+\frac{k q(-q)}{A B}\right] } \\ & -\left[\frac{k q(Q)}{A C}+\frac{k(Q)(-q)}{B C}+\frac{k q(-q)}{A B}\right] \\ = & \frac{k q Q}{3 L}-\frac{k q Q}{L}-\frac{k q^2}{2 L} \\ & \quad-\frac{k q Q}{L}+\frac{k Q q}{L}+\frac{k q^2}{2 L} \\ = & \frac{k q Q}{L}\left[\frac{1}{3}-1\right]=\frac{k q Q}{L} \times\left(\frac{-2}{3}\right) \\ = & \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q Q}{L} \cdot\left(\frac{-2}{3}\right)=-\frac{q Q}{6 \pi \varepsilon_0 L} \end{aligned} $

As $E_1=E_2$ and charges are symmetrically placed on $Y$-axis,

Resultant lies along angle bisector of $\mathbf{E}_1$ and $\mathbf{E}_2$. i.e., along negative $X$-axis.

Using $\left|\frac{d \mathbf{V}}{d r}\right|=|\mathbf{E}|$,

Distance between two equipotentials of a charged plate is

$ \begin{aligned} r & =\frac{\Delta V}{E}=\frac{\Delta V}{\sigma / 2 \varepsilon_0} \\ & =\frac{\Delta V \cdot 2 \varepsilon_0}{\sigma} \\ & =\frac{\Delta V \cdot 4 \pi \varepsilon_0}{2 \pi \cdot \sigma} \end{aligned} $

Here, $\Delta V=90 \mathrm{~V}$,

$ \begin{aligned} \frac{1}{4 \pi \varepsilon_0} & =9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2} \\ \sigma & =2 \times 10^{-7} \mathrm{Cm}^{-2} \end{aligned} $

$ \begin{aligned}So,\,\, r & =\left(\frac{90}{2 \times \pi \times 2 \times 10^{-7} \times 9 \times 10^9}\right) \\ & =\frac{25}{\pi} \times 10^{-3} \mathrm{~m}=\frac{25}{\pi} \mathrm{~mm} \end{aligned} $

Due to field, a proton experiences a force in the direction of field.

To move proton against the field an external agency is required to provide energy to the proton.