Past NEET Entrance Papers

Hint

(d) To find the surface charge density of a charged metal cube, we need to determine the charge per unit area. The surface charge density, denoted by $\sigma$, is given by:
$\sigma=\frac{Q}{A}$
where:
$Q$ is the total charge on the cube
$A$ is the total surface area of the cube
The cube has a side length of 5 cm , so each side of the cube is 5 cm . Since 1 $\mathrm{cm}=0.01 \mathrm{~m}$, the side length in meters is:
$5 \mathrm{~cm}=0.05 \mathrm{~m}$
The cube has 6 faces, and the area of one face is given by:
Area of one face $=(0.05 \mathrm{~m})^2=0.0025 \mathrm{~m}^2$
Therefore, the total surface area of the cube is:
$A=6 \times 0.0025 \mathrm{~m}^2=0.015 \mathrm{~m}^2$
The total charge, $Q$, is given as $6 \mu \mathrm{C}$. Converting this to Coulombs:
$6 \mu \mathrm{C}=6 \times 10^{-6} \mathrm{C}$
Now, plugging the values into the formula for surface charge density:
$\sigma=\frac{6 \times 10^{-6} \mathrm{C}}{0.015 \mathrm{~m}^2}=4 \times 10^{-4} \mathrm{Cm}^{-2}$
Therefore, the surface charge density on the cube is:
$\sigma=0.4 \times 10^{-3} \mathrm{Cm}^{-2}$

Hint

(d) The electric potential $(\mathrm{V})$ at a distance $(\mathrm{r})$ from a point charge $(\mathrm{q})$ is given by:
$V=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$
where $\frac{1}{4 \pi \varepsilon_0}$ is Coulomb’s constant.
In this case, we have:
$\begin{aligned} & q=4 \times 10^{-7} \mathrm{C} \\ & r=9 \mathrm{~cm}=0.09 \mathrm{~m} \\ & \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \end{aligned}$
Substituting these values into the equation for electric potential, we get:
$\begin{aligned} & V=\left(9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\right) \frac{\left(4 \times 10^{-7} \mathrm{C}\right)}{(0.09 \mathrm{~m})} \\ & V=4 \times 10^4 \mathrm{~V} \end{aligned}$
Therefore, the value of electric potential at a distance of 9 cm from the point charge $4 \times 10^{-7} \mathrm{C}$ is $4 \times 10^4 \mathrm{~V}$.

Hint

(d) A charged spherical shell has the same potential at all points inside the shell and on its surface. This property arises because the electric field inside the shell is zero.
The potential $V$ at any point inside the shell (including the center $C$ ) and on the surface $P$ is given by:
$V=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R}$
where
$\begin{aligned} & q=1 \mu C=1 \times 10^{-6} C \\ & R=3 \mathrm{~cm}=0.03 \mathrm{~m}, \text { and } \\ & \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text { SIunits. } \end{aligned}$
Since the potential $V$ is the same at both points $C$ and $P$. the potential difference $\Delta V=V_P-V_C=0$.
Thus, the potential difference between $C$ and $P$ is Zero.

Hint

(c) The potential $V$ at any point, at distance $r$ from centre of dipole $=\frac{K P \cos \theta}{r^2}$
At axial point where $\theta=0^{\circ}, V=\frac{K P}{r^2}=\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2}=9 \times 10^3 \mathrm{~V}$
At axial point where $\theta=180^{\circ}, V=\frac{-K P}{r^2}=-9 \times 10^3 \mathrm{~V}$

Hint

(d) How to solve?
Find the electric field due to each sheet at point $P$, then add them vectorially.
Step 1: Find the electric field due to sheet $\boldsymbol{A}$ at point $\boldsymbol{P}$.
$\begin{aligned} E_A & =\frac{\sigma}{2 \varepsilon_0} \\ E_A & =\frac{\varepsilon_0}{2 \varepsilon_0} \\ E_A & =\frac{1}{2} N / C \end{aligned}$
The electric field due to sheet $A$ is directed away from the sheet.
Step 2: Find the electric field due to sheet $\boldsymbol{B}$ at point $\boldsymbol{P}$.
$\begin{aligned} E_B & =\frac{\sigma}{2 \varepsilon_0} \\ E_B & =\frac{\varepsilon_0}{2 \varepsilon_0} \\ E_B & =\frac{1}{2} N / C \end{aligned}$
The electric field due to sheet $\boldsymbol{B}$ is directed towards the sheet.
Step 3: Add the electric fields due to each sheet vectorially.
Since the electric fields due to each sheet are equal in magnitude and opposite in direction, they cancel each other out.
$\begin{aligned} & E_P=E_A+E_B \\ & E_P=\frac{1}{2}-\frac{1}{2} \\ & E_P=0 \mathrm{~N} / \mathrm{C} \end{aligned}$
The electric field at the midpoint $P$ is $0 \mathrm{~N} / \mathrm{C}$.

Hint

(c) A-IV, B-III, C-II, D-I

Hint

(b) Initial velocity of the particle $=u=0$
let, at $t$ velocity of the particle $=v$ and acceleration $=a$. using formula, $\mathrm{v}=\mathrm{u}+\mathrm{at}$, we get, $\mathrm{v}=\mathrm{at}=\frac{\mathrm{F}{\mathrm{t}}}{\mathrm{m}}$
In electrostatic, $\mathrm{F}=\mathrm{qE}$ so, $\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}} \mathrm{t}$
The kinetic energy of the particle, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{qE}}{\mathrm{m}} \mathrm{t}\right)^2=\frac{\mathrm{q}^2 \mathrm{E}^2 \mathrm{t}^2}{2 \mathrm{~m}}$

Hint

(d)

$\begin{aligned} & \sigma=\frac{Q}{6 \times A}=\frac{6 \times 10^{-6}}{6 \times 5 \times 10^{-2} \times 5 \times 10^{-2}} \\ & \text { Where }\left[A=(\text { side })^2\right] \\ & \sigma=\frac{10^{-6} \times 10^4}{25}=0.04 \times 10^{-2} \mathrm{C} / \mathrm{m}^2 \end{aligned}$

Hint

(b)

$\begin{aligned} &\phi=\frac{\mathrm{q}_{\text {inside }}}{\varepsilon_0}\\ &\text { only depends on charge enclosed by surface. } \end{aligned}$

Hint

(d) The Gaussian surface for the charge placed at the center of the cube will spread out equally through all sides of the cube. According to Gauss’s Law, electric flux $\Phi$ through a closed surface is equal to the charge enclosed $Q$ divided by the permittivity of free space $\epsilon_0$ :
$\Phi_{\text {total }}=\frac{Q}{\epsilon_0}$
Given that the cube has 6 faces, and due to symmetry, the flux through each face will be equal, the flux through any one face $\Phi_{\text {face }}$ is:
$\Phi_{\text {face }}=\frac{\Phi_{\text {itatal }}}{6}=\frac{Q}{6 \epsilon_0}$
Since the charge is given in microcoulombs ( $\mu \mathrm{C}$ ), we need to convert it to coulombs by multiplying with $10^{-6}$.
$\Phi_{\mathrm{face}}=\frac{Q \times 10^{-6}}{6 \epsilon_0}$
This matches option $D$, which means the correct flux through one face of the cube given a charge $Q$ microcoulombs at the center is $\frac{Q}{6 \epsilon_0} \times 10^{-6}$.

Hint

(c) 1. Understand the Concept of Electric Field and Potential:
The electric potential $V$ at the surface of a charged conducting sphere is given by the formula:
$V=\frac{k Q}{R}$
where $k$ is Coulomb’s constant, $Q$ is the charge on the sphere, and $R$ is the radius of the sphere.
2. Relate Charge to Potential:
From the above formula, we can express the charge $Q$ in terms of the potential $V$ :
$Q=\frac{V R}{k}$
3. Electric Field Outside the Sphere:
For a conducting sphere, the electric field $E$ at a distance $r$ from the center (where $r>R$ ) can be calculated using the formula for the electric field due to a point charge:
$E=\frac{k Q}{r^2}$
4. Substituting Charge into Electric Field Formula:
Now substitute the expression for $Q$ into the electric field formula:
$E=\frac{k\left(\frac{V R}{k}\right)}{r^2}$
Simplifying this gives:
$E=\frac{V R}{r^2}$
5. Final Expression for Electric Field:
Therefore, the electric field $E$ at a distance $r$ from the center of the sphere (where $r>R$ ) is:
$E=\frac{V}{r^2} \cdot R$

Hint

(c) The torque $\tau$ experienced by an electric dipole in an electric field is given by the formula:
$\tau=p E \sin \theta$
where $p$ is the electric dipole moment, $E$ is the electric field intensity, and $\theta$ is the angle between the dipole and the electric field. The electric dipole moment p can be expressed as:
$p=q d$
where q is the charge on the dipole, and d is the dipole length.
We are given the following values:
Torque $\tau=4 \mathrm{~N} \cdot \mathrm{~m}$
Electric field intensity $\mathrm{E}=2 \times 10^5 \mathrm{NC}^{-1}$
Angle $\theta=30^{\circ}$
Dipole length $\mathrm{d}=2 \mathrm{~cm}=0.02 \mathrm{~m}$
We need to find the charge $q$ on the dipole. Let’s first solve for the electric dipole moment p :
$\begin{aligned} & \tau=p E \sin \theta \\ & \Rightarrow p=\frac{\tau}{E \sin \theta} \end{aligned}$
Substituting the given values:
$p=\frac{4}{\left(2 \times 10^5\right) \sin 30^{\circ}}=\frac{4}{\left(2 \times 10^5\right)(0.5)}=\frac{4}{10^5}=4 \times 10^{-5} \mathrm{C} \mathrm{~m}$
Now, let’s solve for the charge qusing the formula:
$\begin{aligned} & \Rightarrow p=q d \\ & q=\frac{p}{d} \end{aligned}$
Substituting the values for $p$ and $d$ :
$q=\frac{4 \times 10^{-5}}{0.02}=2 \times 10^{-3} \mathrm{C}=2 \mathrm{mC}$

Hint

(d) We know electric flux is proportional to the number of electric field lines and the term $\oint_s$ E.dS represents electric flux over the closed surface.
It means $\oint_s$ E.dS represents the algebraic sum of the number of flux lines entering the surface and number of flux lines leaving the surface.
If $\oint_s E . d S=0$, this means the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
From Gauss’ law, we know $\oint_s E . d S=\frac{q}{\epsilon_0}$, here q is the charge enclosed by , the closed surface. If $\oint_s E . d S=0$ then $\mathrm{q}=0$, i.e., net charge enclosed by the surface must be zero. Hence all other charges must necessarily be outside the surface. This is because of the fact that charges outside the surface do not contribute to the electric flux.

Alternate:

$\begin{aligned} &\phi_{\text {closed }}=0\\ &\text { So, } \phi_{\text {in }}=\phi_{\text {out }} \end{aligned}$
Number of field lines entering is equal number of field lines leaving.

Hint

(d)

$\begin{aligned} & \mathrm{v}=\frac{\mathrm{Kq}}{2 \times 10^{-2}}-\frac{\mathrm{Kq}}{8 \times 10^{-2}} \\ & =\mathrm{Kq}\left[\frac{3}{8}\right] \times 10^2 \end{aligned}$

Hint

(a) Work to bring a charge ‘ $q$ ‘ from infinity to a point is $\mathrm{W}=\mathrm{q} \mathrm{V}$.
Where $V$ is the potential at that point.
$\text { Work done }=q_0 \cdot\left(V_0-V_{\infty}\right)$
Here, potential at the centre of the hexagon is due to the charges present at its corner.

$\begin{aligned} & =\left\{\frac{3 k q}{d}+\left(\frac{-3 k q}{d}\right)\right\} q_0 \\ & =\text { Zero } \end{aligned}$
where ‘ $d$ ‘ is the distance of the corner to centre (as six equilateral triangles are formed when we joined the corners to centre in a hezagon).
So, $V=0$ and $W=0$
So, no work is done to bring a charge from infinity to the centre of the given hexagon.

Hint

(c)

$\begin{aligned} &\begin{aligned} & \mathrm{dV}=-\vec{E} \cdot \mathrm{~d} \vec{r} \\ & \mathrm{dV}=-\mathrm{Edr} \cos \theta \end{aligned}\\ &\text { For equipotential surface, }\\ &\mathrm{dV}=0\\ &\cos \theta=0\\ &\Rightarrow \theta=90^{\circ} \end{aligned}$

Hint

(b) Potential of conducting hollow sphere $=\frac{K Q}{R}$
Now, Q = same
$\Rightarrow V \propto \frac{1}{R} \Rightarrow$ more the radius less will be the potential.
$\Rightarrow$ Hence potential would be more on smaller sphere.

Hint

(b)

How to solve?
Use the formula for the electric field intensity due to an electric dipole to find the variation of the electric field intensity with distance.

Step 1: Identify the system as an electric dipole.
The two point charges $-q$ and $+q$ separated by a distance $L$ form an electric dipole.

Step 2: Determine the electric dipole moment.
$p=q L$
Step 3: Use the formula for the electric field intensity due to an electric dipole to find the variation of the electric field intensity with distance.
Since $R \gg L$, we can consider the point of observation to be far away from the dipole. In this case, the electric field intensity due to an electric dipole varies as $\frac{1}{r^3}$.
$E=\frac{2 p}{4 \pi \varepsilon_0 R^3}$
$E \propto \frac{1}{R_3}$
The magnitude of electric field intensity at a distance $R(R \gg L)$ varies as $\frac{1}{R^3}$.

Hint

(c) 

$\begin{aligned} &F=\frac{k q_1 q_2}{r^2}\\ &\text { Where } \mathrm{k} \text { is a constant of proportionality and has the value } 9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \text {. } \end{aligned}$
The electrostatic force experienced by charge q due to Q is $F=\frac{k q Q}{R^2}$
Since there are $\mathbf{1 2}$ charges of the same magnitude, the electrostatic force experienced by the charge q due to each of the 12 charges can be named $F_1, F_2, F_3, F_4, \ldots . F_{12}$.
Therefore, the net force on charge $q$,
$\mathbf{F}^{\prime}=\mathrm{F}_1+\mathrm{F}_2+\mathrm{F}_3+\mathrm{F}_4+\mathrm{F}_5+\mathrm{F}_6+\mathrm{F}_7+\mathrm{F}_8+\mathrm{F}_9+\mathrm{F}_{10}+\mathrm{F}_{11}+\mathrm{F}_{12}$
The forces acting on charge $q$ are as shown. Since they all are equal and opposite, the net force is zero.
$\Rightarrow F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}+F_{11}+F_{12}=F^{\prime}=0 \dots(1)$
When one of the 12 charges is removed, one of the forces acting on the charge $q$ vanishes.
If $\mathrm{F}_1$ is the force that vanishes due to the removal of a charge, it will be subtracted from the net force.
Subtracting $F_1$ on both LHS and RHS in (1),
$\begin{aligned} & \left(F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}+F_{11}+F_{12}\right)-F_1=\mathbf{0}-\mathbf{F}_{\mathbf{1}}=\mathbf{F}^{\prime} \\ & \Rightarrow \text { Net force, } \mathbf{F}^{\prime}=-\mathbf{F}_1 \end{aligned}$
Therefore, the net force is equal to $\frac{k q Q}{R^2}$ and will be directed towards the position of the removed charge.

Hint

(c) How to solve?
Find the ratio of the electrostatic force to the gravitational force, then substitute the given values to find the ratio of $k$ to $\boldsymbol{G}$.
Step 1: Find the ratio of the electrostatic force to the gravitational force.
$\begin{aligned} & \frac{F_e}{F_g}=\frac{k \frac{q_1 g_2}{r^2}}{G \frac{m_1 m_2}{r^2}} \\ & \frac{F_e}{F_g}=\frac{k q_1 q_2}{G m_1 m_2} \end{aligned}$
Step 2: Substitute the given values into the equation.
$\begin{aligned} & \frac{F_e}{F_g}=\frac{k\left(1.6 \cdot 10^{-19}\right)^2}{G\left(9.11 \cdot 10^{-31}\right)\left(1.67 \cdot 10^{-27}\right)} \\ & 2.4 \cdot 10^{39}=\frac{k\left(1.6 \cdot 10^{-19}\right)^2}{G\left(9.11 \cdot 10^{-31}\right)\left(1.67 \cdot 10^{-27}\right)} \end{aligned}$
Step 3: Solve for $\frac{k}{G}$.
The ratio of the proportionality constant $k$ to the gravitational constant $\boldsymbol{G}$ is nearly $10^{40}$.

Hint

(b) Polar molecules have centres of positive and negative charges separated by some distance, so they have permanent dipole moment.

Hint

(c) The potential energy of electric dipole in an external electric field is written as
$U=-\vec{P} \cdot \vec{E} \dots(1)$
Where P is the dipole and E is the electric field.
Using equation (1) we get;
$\begin{aligned} & U=-\vec{P} \cdot \vec{E} \\ & \Rightarrow U=-\mathrm{PE} \cos \theta \end{aligned}$
The angle between the electric field and the electric dipole is $180^{\circ}$, therefore,
$\begin{aligned} & U=-P E \cos 180^{\circ} \\ & U=+P E \end{aligned}$
On moving towards the right electric field strength decrease therefore potential energy decrease.
The net force on the electric dipole is towards the right and the net torque acting on it is zero.
So, it will move towards the right.

Hint

(c)

$\begin{aligned} & Q_1=\frac{\sum Q}{R_1+R_2} \times R_1 \\ & Q_2=\frac{\sum Q}{R_1+R_2} \times R_2 \\ & \sigma_1=\frac{Q_1}{4 \pi R_1^2}=\frac{\sum Q}{R_1+R_2} \times \frac{R_1}{4 \pi R_1^2} \propto \frac{1}{R_1} \\ & \sigma_2=\frac{Q_2}{4 \pi R_2^2}=\frac{\sum Q}{R_1+R_2} \times \frac{R_2}{4 \pi R_2^2} \propto \frac{1}{R_2} \\ & \frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1} \end{aligned}$

Hint

Let charge and radius of smaller drop is $q$ and $r$ respectively
For smaller drop, $\mathrm{V}=\frac{k q}{r}=220 \mathrm{~V}$
Let R be radius of bigger drop,
As volume remains the same
$\begin{aligned} & \left(\frac{4}{3} \pi r^3\right) \times 27=\frac{4}{3} \pi R^3 \\ & \Rightarrow \mathrm{R}=3 \mathrm{r} \end{aligned}$
Now, using charge conservation,
$\begin{aligned} & \mathrm{Q}=27 \mathrm{q} \\ & \mathrm{~V}_{\text {big drop }}=\frac{k Q}{r}=\frac{k(27 q)}{r}=9\left(\frac{k q}{r}\right) \\ & =9 \times 220=1980 \mathrm{~V} \end{aligned}$

Hint

(a)

$\begin{aligned} &\text { Electric potential due to electric dipole (V) }\\ &\begin{aligned} & =\frac{1}{4 \pi \varepsilon_0} \frac{p \cdot \cos \theta}{r^2} \\ & V=\frac{9 \times 10^9 \times 16 \times 10^9 \times \cos 60^{\circ}}{0.36} \end{aligned}\\ &\mathrm{V}=200 \mathrm{~V} \end{aligned}$

Hint

(d) Potential V is constant. Electric field E is the differentiation of potential V with respect to distance. Differentiation of a constant is zero. So $\mathrm{E}=0$.

Hint

(a) 

$\begin{aligned} &\text { The electric field for the conducting sphere, away from the surface is given by, }\\ &\begin{aligned} & E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \\ & =\frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{225 \times 10^{-4}} \\ & =0.128 \times 10^6 \\ & =1.28 \times 10^5 \mathrm{~N} / \mathrm{C} \end{aligned} \end{aligned}$

Hint

(d) 1. Calculating the Potential: The electric potential $V$ at point P due to the dipole can be expressed as:
$V=\frac{k p \cos \theta}{r^2}$
Here, $k$ is the Coulomb’s constant, $p$ is the dipole moment, and $\theta$ is the angle with respect to the dipole axis.
2. Finding the Electric Field: The electric field $E$ is related to the potential by the relation:
$E=-\frac{d V}{d r}$
For the radial component $E_r$ :
$E_r=-\frac{d V}{d r}=-\frac{d}{d r}\left(\frac{k p \cos \theta}{r^2}\right)$
After differentiating, we find:
$E_r=\frac{2 k p \cos \theta}{r^3}$
3. Calculating the Transverse Component:
For the transverse component $E_\theta$ :
$E_\theta=-\frac{1}{r} \frac{d V}{d \theta}$
After differentiating, we find:
$E_\theta=\frac{k p \sin \theta}{r^3}$
4. Resultant Electric Field:
The resultant electric field $E$ at the equatorial point is given by:
$E=\sqrt{E_r^2+E_\theta^2}$
At the equatorial plane, $\theta=90^{\circ}$ implies $\cos \left(90^{\circ}\right)=0$ and $\sin \left(90^{\circ}\right)=1$. Therefore:
$E_r=0 \quad \text { and } \quad E_\theta=\frac{k p}{r^3}$
Thus, the resultant electric field at the equatorial point is:
$E=\frac{k p}{r^3}$
5. Expressing in Vector Form:
The constant $k$ can be expressed as $k=\frac{1}{4 \pi \epsilon_0}$, leading to:
$\vec{E}=-\frac{1}{4 \pi \epsilon_0} \frac{\vec{p}}{r^3}$

Hint

(c) Force due to mutual attraction between the electron and proton. (when, $r=1.6 Å =1.6 \times 10^{-10} \mathrm{~m}$ ) is given as
$\begin{aligned} & F=9 \times 10^9 \times \frac{e^2}{r^2} \\ & =9 \times 10^9 \times \frac{\left(1.6 \times 10^{-19}\right)^2}{\left(1.6 \times 10^{-10}\right)^2}=9 \times 10^{-9} \mathrm{~N} \end{aligned}$
$\therefore$ Acceleration of electron
$=\frac{F}{m_e}=\frac{9 \times 10^{-9}}{9 \times 10^{-31}}=10^{22} \mathrm{~m} / \mathrm{s}^2$

Hint

(b) The electric flux is written as;
$\phi=\mathrm{E} . \mathrm{A}$
Here we have $E$ as the electric field and $A$ is the area.
and Electric field is written as;
$\vec{E}=\frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}$
Here $Q$ is the charge, and $r$ is the distance.
Calculation:
Here we are given that the sphere encloses an electric dipole with charges.
When the sphere is enclosed the electric flux coming outside and the electric field outside the sphere or electric flux outside the sphere is zero.
Hence, the total electric flux across the sphere is zero.

Hint

(a) Charge Q will be distributed over the surface of hollow metal sphere.
(i) For $\mathrm{r}<\mathrm{R}$ (inside) [At a point inside the hollow sphere]

By Gauss’s law, $\oint \overrightarrow{\mathrm{E}}_{\mathrm{in}} \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{q}_{\mathrm{en}}}{\varepsilon_0}=0$
As enclosed charge is $=0$
So, $\mathrm{E}_{\text {in }}=0$ the electric field inside the hollow sphere is always zero.
(ii) For $r>R$ (outside)
[At a point outside hollow sphere]

$\text { By Gauss’s law, } \oint \overrightarrow{\mathrm{E}}_0 \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{q}_{\mathrm{en}}}{\varepsilon_0} \quad\left(\mathrm{q}_{\mathrm{en}}=\mathrm{Q}\right)$
$\begin{aligned} \therefore & \mathrm{E}_0 4 \pi \mathrm{r}^2=\frac{\mathrm{Q}}{\varepsilon_0} \\ & \therefore \mathrm{E}_0 \propto \frac{1}{\mathrm{r}^2} \end{aligned}$

Hint

(b) Initially $\mathrm{F}=-\frac{K Q^2}{r^2}$

If $25 \%$ of charge of $A$ transferred to $B$ then after the transformation of charge.
$\mathrm{q}_{\mathrm{A}}=Q-\frac{Q}{4}=\frac{3 Q}{4}$
and $\mathrm{q}_{\mathrm{b}}=-\mathrm{Q}+\frac{Q}{4}=-\frac{3 Q}{4}$
New force, $\mathrm{F}_1=\frac{K\left(\frac{3 Q}{4}\right)\left(-\frac{3 Q}{4}\right)}{r^2}$
$\begin{aligned} & =-\frac{9}{16} \frac{K Q^2}{r^2} \\ & =\frac{9 F}{16} \end{aligned}$

Hint

(c)

$\begin{aligned} &\text { (c) Electric field due to line charge (1) }\\ &\overrightarrow{\mathrm{E}}_1=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{R}} \hat{\mathrm{i}} / \mathrm{C} \end{aligned}$
Electric field due to line charge (2)
$\begin{aligned} & \overrightarrow{\mathrm{E}}_2=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{R}} \hat{\mathrm{i}} \mathrm{~N} / \mathrm{C} \\ & \overrightarrow{\mathrm{E}}_{\text {net }}=\overrightarrow{\mathrm{E}}_1+\overrightarrow{\mathrm{E}}_2 \\ & =\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{R}} \hat{\mathrm{i}}+\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{R}} \hat{\mathrm{i}} \\ & =\frac{\lambda}{\pi \varepsilon_0 \mathrm{R}} \hat{\mathrm{i}} / \mathrm{C} \end{aligned}$

Note: Both $\overrightarrow{\mathrm{E}}_1$ and $\overrightarrow{\mathrm{E}}_2$ are in the same direction. Electric field at a point due to +ve charge $(+q)$ acts away from the charge and due to negative charge $(-q)$ it acts towards the charge.

Hint

(a) Force experienced by a charged particle in an electric field, $\mathrm{F}=\mathrm{qE}$
As $\mathrm{F}=\mathrm{ma}$
$\therefore m a=q E$
$\Rightarrow \mathrm{a}=\frac{q E}{m}$
As electron and proton both fall from same height at rest. Then initial velocity $=0$
From the formula, $s=u t+\frac{1}{2} a t^2$
$\begin{aligned} & \therefore \mathrm{h}=\frac{1}{2} a t^2 \\ & \Rightarrow \mathrm{~h}=\frac{1}{2} \frac{q E}{m} t^2 \\ & \therefore \mathrm{t}=\sqrt{\frac{2 h m}{q E}} \end{aligned}$
$\Rightarrow \mathrm{t} \propto \sqrt{m}$ as ‘ $q$ ‘ is same for electron and proton.
Since, electron has smaller mass so it will take smaller time.

Hint

(b)

$\text { Acceleration, } \mathrm{a}=\frac{v-u}{t}=\frac{6-0}{1}=6 \mathrm{~ms}^{-2}$
$\begin{aligned} &\text { For } t=0 \text { to } t=1 \mathrm{~s} \text {, }\\ &\mathrm{S}_1=\frac{1}{2} \times 6(1)^2=3 \mathrm{~m} \dots(i) \end{aligned}$
$\begin{aligned} &\text { For } t=1 \mathrm{~s}=\text { to } \mathrm{t}=2 \mathrm{~s} \text {, }\\ &\mathrm{S}_2=6.1-\frac{1}{2} \times 6(1)^2=3 \mathrm{~m} \dots(ii) \end{aligned}$
Total displacement $\mathrm{S}=\mathrm{S}_1+\mathrm{S}_2+\mathrm{S}_3=3 \mathrm{~m}$
Average velocity $=\frac{3}{3}=1 \mathrm{~ms}^{-1}$
Total distance travelled $=9 \mathrm{~m}$
Average speed $=\frac{9}{3}=3 \mathrm{~ms}^{-1}$

Hint

(b) Net charge on one Hatom $=-\mathrm{e}+(\mathrm{e}+\Delta \mathrm{e})=\Delta \mathrm{e}$ According to question, the net electrostatic force $\left(\mathrm{F}_{\mathrm{E}}\right)=$ gravitational force $\left(\mathrm{F}_{\mathrm{G}}\right)$
$\mathrm{F}_{\mathrm{E}}=\mathrm{F}_{\mathrm{G}}$
$\begin{aligned} &\begin{aligned} & \Rightarrow \frac{1}{4 \pi \varepsilon_0} \frac{\Delta e^2}{d^2}=\frac{G m^2}{d^2} \\ & \Rightarrow \Delta e=m \sqrt{\frac{G}{K}}\left(\frac{1}{4 \pi \varepsilon_0}=k=9 \times 10^9\right) \\ & =1.67 \times 10^{-27} \sqrt{\frac{6.67 \times 10^{-11}}{9 \times 10^9}} \end{aligned}\\ &\Delta e \approx 1.436 \times 10^{-37} C \end{aligned}$

Hint

(a) As the regions are of equipotentials, so Work done $\mathrm{W}=\mathrm{q} \Delta \mathrm{V}$.
Since $\Delta \mathrm{V}$ is the same in all the cases hence work done will also be the same in all the cases.

Hint

(b)

From figure $\mathrm{T} \cos \theta=\mathrm{mg}$
$\mathrm{T} \sin \theta=\frac{\mathrm{kq}^2}{\mathrm{x}^2}$
$\begin{aligned} \text { so, } \quad \tan \theta & =\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{mg}} \simeq \theta \\ \tan \theta & =\frac{\mathrm{kq}^2}{\mathrm{x}^2 \mathrm{mg}} \end{aligned}$
$\begin{aligned} &\text { As } \theta \text { is small } \tan \theta \approx \sin \theta=\frac{x}{2 \ell}\\ &\therefore \quad \frac{\mathrm{kq}^2}{\mathrm{x}^2 \mathrm{mg}}=\frac{\mathrm{x}}{2 \ell} \end{aligned}$
or $\quad x^3 \propto q^2 \quad \ldots(1)$
or $\quad x^{3 / 2} \propto q \ldots(2)$
Differentiating eq. (1) w.r.t. time
$3 x^2 \frac{d x}{d t} \propto 2 q \frac{d q}{d t}$ but $\frac{d q}{d t}$ is constant
so $\mathrm{x}^2(\mathrm{v}) \propto \mathrm{q}$
Replace $q$ from eq. (2)
$x^2(v) \propto x^{3 / 2} \quad \text { or } \quad v \propto x^{-1 / 2}$

Hint

(b)

$\begin{aligned} &\text { Torque is given by } \tau=p E \sin \theta\\ &\tau=p E \sin \theta=q l E \sin \theta\\ &\Rightarrow q=\tau / l E \sin \theta\\ &=4 /\left(2 \times 10^{-2} \times 0.5 \times 2 \times 10^5\right)=2 \mathrm{mC} \end{aligned}$

Hint

(c) According to question, electric field varies as
$\mathrm{E}=\mathrm{Ar}$
Here $r$ is the radial distance.
At $\mathrm{r}=\mathrm{a}, \mathrm{E}=\mathrm{Aa} \dots(i)$
Net flux emitted from a spherical surface of radius a is $\phi_{n e t}=\frac{q_{e n}}{\varepsilon_0}$
$\begin{aligned} & \Rightarrow \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=(A a) \times\left(4 \pi a^2\right)=\frac{q_{e n}}{\varepsilon_0}[\text { Using equation (i)] } \\ & \therefore q_{e n}=4 \pi \varepsilon_0 A a^3 \end{aligned}$

Hint

(d) Potential in a region $V=6 x y-y+2 y z$
As we know the relation between electric potential and electric field is $\vec{E}=\frac{-d V}{d x}$
$\begin{aligned} & \vec{E}=\left(\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \widehat{k}\right) \\ & \vec{E}=[(6 y \hat{i}+(6 x-1+2 z) \hat{j}+(2 y) \widehat{k})] \\ & \vec{E}_{(1,1,0)}=-(6 \hat{i}+5 \hat{j}+2 \widehat{k}) \end{aligned}$

Hint

(b) For the conducting sphere, Potential at the centre $=$ Potential on the sphere
$=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$
Electric field at the centre $=0$

Hint

(d) 

$\begin{aligned} & \vec{E}=-\frac{\partial V}{\partial x} \hat{i}-\frac{\partial V}{\partial y} \hat{j}-\frac{\partial V}{\partial z} \widehat{k} \\ & =-[(6-8 y) \hat{i}+(-8 x-8+6 z) \hat{j}+(6 y) \hat{k}] \\ & \text { At }(1,1,1), \vec{E}=2 \hat{i}+10 \hat{j}-6 \hat{k} \\ & \Rightarrow(\vec{E})=\sqrt{2^2+10^2+6^2}=\sqrt{140}=2 \sqrt{35} \\ & \therefore F=q \vec{E}=2 \times 2 \sqrt{35}=4 \sqrt{35} N \end{aligned}$

Hint

(b) When electric dipole is aligned parallel $\theta=0^{\circ}$ and the dipole is rotated by $90^{\circ}$ i.e., $\theta=90^{\circ}$.
Energy required to rotate the dipole
$\begin{aligned} W & =U_f-U_i=\left(-p E \cos 90^{\circ}\right)-\left(-p E \cos 0^{\circ}\right) \\ & =p E \end{aligned}$

Hint

(a) The system of three charges will be in equilibrium.

For this, force between charge at $A$ and $B$ + force between charge at point $O$ and either at $A$ or $B$ is zero.
$\text { i.e., } \frac{K Q^2}{(2 r)^2}+\frac{K Q q}{(r)^2}=0$
By solving we get,
$q=-\frac{Q}{4}$

Hint

(d)

$T \cos \theta=m g$
$\mathrm{T} \sin \theta=\mathrm{F}$
$\therefore \quad \tan \theta=\frac{\mathrm{F}}{\mathrm{mg}}$
From figure, $\tan \theta=\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{mg}}$
$\begin{aligned} & \Rightarrow \frac{\mathrm{r} / 2}{\mathrm{y}}=\frac{\frac{\mathrm{kq}^2}{\mathrm{r}^2}}{\mathrm{mg}} \\ & {\left[\because \mathrm{~F}=\frac{\mathrm{kq}^2}{\mathrm{r}^2} \text { from coulomb’s law }\right]} \end{aligned}$
$\Rightarrow \mathrm{r}^3 \propto \mathrm{y}$
When string is clamped at half the height
$\mathrm{r}^{\prime 3} \propto \frac{\mathrm{y}}{2}$
Dividing the above two equations, we have
$\frac{\mathrm{r}^{\prime}}{\mathrm{r}}=\frac{1}{2^{1 / 3}} \Rightarrow \mathrm{r}^{\prime}=\frac{\mathrm{r}}{\sqrt[3]{2}}$

Hint

(d) In the direction of electric field, electric potential decreases.
$\therefore \mathrm{V}_{\mathrm{B}}>\mathrm{V}_{\mathrm{C}}>\mathrm{V}_{\mathrm{A}}$

Hint

(b)

$Q_{\text {total }}=5 \times 10^{-2} \mathrm{C}-1 \times 10^{-2} \mathrm{C}=4 \times 10^{-2} \mathrm{C}$
We know that, Charge distribution $\propto$ radius
$\begin{aligned} & \therefore Q_{\text {small }}: Q_{b i g}=1: 3 \\ & Q_{b i g}=\frac{3}{4} \times 4 \times 10^{-2}=3 \times 10^{-2} \mathrm{C} \end{aligned}$

Hint

(a)

Let the distance of the center (O) from each corner be $r$, then
$\begin{aligned} &\text { The total potential at the point is given by }\\ &\begin{aligned} & \Rightarrow V_{\text {total }}=\mathrm{V}_1+\mathrm{V}_2+\mathrm{V}_3+\mathrm{V}_4 \\ & \Rightarrow V_{\text {total }}=-\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}+\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{r}-\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}+\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{r} \\ & \Rightarrow V_{\text {total }}=\frac{1}{4 \pi \varepsilon_o r}(-q+2 q-Q+2 Q) \\ & \Rightarrow V_{\text {total }}=\frac{1}{4 \pi \varepsilon_0 r}(q+Q) \end{aligned} \end{aligned}$
$\begin{aligned} &\text { According to question, } V_{\text {total }}=0 \text {, then the above equation becomes }\\ &\begin{aligned} & \Rightarrow 0=\frac{1}{4 \pi \epsilon_o r}(q+Q) \\ & \Rightarrow q+Q=0 \\ & \Rightarrow \mathrm{Q}=-\mathrm{q} \end{aligned} \end{aligned}$

Hint

(b)

Eight identical cubes are required so that the given charge $q$ appears at the centre of the bigger cube.
Thus, the electric flux passing through the given cube is
$\phi=\frac{1}{8}\left(\frac{q}{\varepsilon_0}\right)=\frac{q}{8 \varepsilon_0}$

Hint

(a) The torque on the dipole is given as $\tau=p E \sin \theta$
The potential energy of the dipole in the electric field is given as $U=-p E \cos \theta$

Hint

(c) According to Gauss’s law
$\phi_E=\frac{Q_{\text {enclosed }}}{\varepsilon_0}$
If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same. This is because electric flux depends only on the charge enclosed by the surface.

Hint

(d) According to figure, $A C=B C$
Potential at $D=$ potential at $E$
$\begin{aligned} &V_D=V_E\\ &\text { We have, } W=Q\left(V_E-V_D\right)\\ &\Rightarrow \mathrm{W}=0 \end{aligned}$

Hint

(a)

$\vec{E}=-\vec{\nabla} V$
$\vec{E}=-\left[\frac{d V}{d x} \hat{i}+\frac{d V}{d y} \hat{j}+\frac{d V}{d z} \hat{k}\right]$
$\begin{aligned} & \text { Given, } \mathrm{V}=4 x^2 \\ & \begin{aligned} \therefore \quad E & =-\hat{i} \frac{d\left(4 x^2\right)}{d x} \\ \quad= & -8 x \hat{i} \text { volt } / \mathrm{meter} \\ \therefore \overrightarrow{\mathrm{E}}_{(1,0,2)} & =-8 \hat{i} \mathrm{~V} / \mathrm{m} \end{aligned} \end{aligned}$
So electric field is along the negative X -axis.

Hint

(c) Distance of point A from the two +q charges $=\mathrm{L}$.
Distance of point A from the two -q charges
$=\sqrt{L^2+(2 L)^2}=\sqrt{5} L$

$\begin{aligned} & \therefore V_A=\left(\frac{K q}{L} \times 2\right)-\left(\frac{K q}{\sqrt{5} L} \times 2\right) \\ & =\frac{2 K q}{L}\left[1-\frac{1}{\sqrt{5}}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}\right) \end{aligned}$

Hint

(b) Electric field at a point inside a charged conducting spherical shell is zero.

Hint

(c) According to Coulomb’s law, the force of repulsion between the two positive ions each of charge $q$, separated by a distance $d$ is given by
$\begin{aligned} & F=\frac{1}{4 \pi \varepsilon_0} \frac{(q)(q)}{d^2} \\ & F=\frac{q^2}{4 \pi \varepsilon_0 d^2} \\ & q^2=4 \pi \varepsilon_0 F d^2 \\ & q=\sqrt{4 \pi \varepsilon_0 F d^2} \ldots \text { (i) } \end{aligned}$
Since, $q=n e$
where, $\mathrm{n}=$ number of electrons missing from each ion $e=$ magnitude of charge on electron
$\begin{aligned} & \therefore n=\frac{q}{e} \\ & n=\frac{\sqrt{4 \pi \varepsilon_0 F d^2}}{e} \text { (Using (i)) } \\ & =\sqrt{\frac{4 \pi \varepsilon_0 F d^2}{e^2}} \end{aligned}$

Hint

(d) Electric flux, $\phi=E A \cos \theta$, where $\theta$ $=$ angle between $E$ and normal to the surface.
$\begin{aligned} & \text { Here } \theta=\frac{\pi}{2} \\ & \Rightarrow \phi=0 \end{aligned}$

Hint

$\begin{aligned} & \text { (d) Energy }=2 \mathrm{eV} \\ & \Rightarrow \mathrm{eV}_{\mathrm{o}}=2 \mathrm{eV} \Rightarrow \mathrm{~V}_{\mathrm{o}}=2 \text { volt } \\ & \mathrm{E}=\frac{\mathrm{V}_0}{\mathrm{~d}}=\frac{2}{4 \times 10^{-8}} \\ & =0.5 \times 10^8=5 \times 10^7 \mathrm{Vm}^{-1} \end{aligned}$

Hint

(d) The electric potential at a point,
$V=-x^2 y-x z^3+4$
The field
$\begin{aligned} & \vec{E}=-\vec{\nabla} V=-\left(\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \widehat{k}\right) \\ & \therefore \vec{E}=\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\widehat{k}\left(3 x z^2\right) \end{aligned}$

Hint

(d)

$\begin{aligned} & V_A=\frac{1}{4 \pi \varepsilon_0}\left\{\frac{q A}{a}+\frac{q B}{b}+\frac{q C}{c}\right\} \\ & =\frac{4 \pi}{4 \pi \varepsilon_0}\left\{\frac{a^2 \sigma}{a}-\frac{b^2 \sigma}{b}+\frac{c^2 \sigma}{c}\right\} \\ & V_A=\frac{1}{\varepsilon_0}\left\{\frac{a^2 \sigma}{a}-\frac{b^2 \sigma}{b}+\frac{c^2 \sigma}{c}\right\} \\ & V_B=\frac{1}{\varepsilon_0}\left\{\frac{a^2 \sigma}{a}-\frac{b^2 \sigma}{b}+\frac{c^2 \sigma}{c}\right\} \\ & V_C=\frac{1}{\varepsilon_0}\left\{\frac{a^2 \sigma}{a}-\frac{b^2 \sigma}{b}+\frac{c^2 \sigma}{c}\right\} \end{aligned}$
Given $\mathrm{c}=\mathrm{a}+\mathrm{b}$.
If $a=a, b=2 a$ and $c=3 a$ for example, as $c>b>a$,
$\begin{aligned} & V_A=\frac{1}{\varepsilon_0}\left\{\frac{a^2 \sigma}{a}-\frac{4 b^2 \sigma}{2 a}+\frac{c^2 \sigma}{c}\right\} \\ & V_B=\frac{1}{\varepsilon_0}\left\{\frac{a^2 \sigma}{2 a}-\frac{4 a^2 \sigma}{2 a}+\frac{c^2 \sigma}{c}\right\} \\ & V_C=\frac{1}{\varepsilon_0}\left\{\frac{a^2 \sigma}{3 a}-\frac{4 a^2 \sigma}{3 a}+\frac{c^2 \sigma}{c}\right\} \end{aligned}$
It can seen by taking out common factors that
$V_A=V_C>V_B \text { i.e., } V_A=V_C \neq V_B$

Hint

(b) By the symmetry of the figure, the electric fields at O due to the portions AC and BD are equal in magnitude and opposite in direction. So, they cancel each other.

Similarly, the field at O due to CD and AKB are equal in magnitude but opposite in direction. Therefore, the electric field at the centre due to the charge on the part ACDB is E along OK .

Hint

(c)

$\begin{aligned} & V=\frac{1}{4 \pi \varepsilon_0} \times \frac{Q}{R} \\ & =\mathrm{Q} \times 10^{11} \mathrm{volt} \quad \ldots \text { (i) } \\ & E=\frac{1}{4 \pi \varepsilon_0} \times \frac{Q}{R^2} \\ & =\mathrm{V} / \mathrm{R}=\mathrm{Q} \times 10^{11} \times 4 \pi \varepsilon_0 \times 10^{11} \quad \text { from } \ldots \text { (i) } \\ & =4 \pi \varepsilon_0 \times \mathrm{Q} \times 10^{22} \mathrm{volt} / \mathrm{m} \end{aligned}$

Hint

(a) Three point charges $+q,-2 q$ and $+q$ are placed at points $B(x=0, y=a, z=0)$, $O(x=0, y=0, z=0)$ and $A(x=a, y=0, z=0)$ The system consists of two dipole moment vectors due to $(+q$ and $-q)$ and again due to $(+q$ and $-q$ ) charges having equal magnitudes qa units – one along $O A$ and other along $O B$. Hence, net dipole moment,
$p_{\text {net }}=\sqrt{(q a)^2+(q a)^2}=\sqrt{2} q a$ along $\overrightarrow{O P}$ at an angle $45^{\circ}$ with positive $X$-axis.

Hint

(a) Since $\phi_{\text {total }}=\phi_A+\phi_B+\phi_C=\frac{q}{\varepsilon_0}$,
where $q$ is the total charge.
As shown in the figure, flux associated with the curved surface B is $\phi=\phi_B$
Let us assume flux linked with the plane surfaces $A$ and $C$ be
$\phi_A=\phi_C=\phi^{\prime}$
Therefore,
$\frac{q}{\varepsilon_0}=2 \phi^{\prime}+\phi_B=2 \phi^{\prime}+\phi$
$\Rightarrow \phi^{\prime}=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$

Hint

(c)

Potential at $\mathrm{C}=\mathrm{V}_{\mathrm{C}}=0$
Potential at $\mathrm{D}=\mathrm{V}_{\mathrm{D}}$
$=k\left(\frac{-q}{L}\right)+\frac{k q}{3 L}=-\frac{2}{3} \frac{k q}{L}$
Potential difference
$\begin{aligned} & \mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}=-\frac{2}{3} \frac{k q}{L}=\frac{1}{4 \pi \varepsilon_0}\left(-\frac{2}{3} \cdot \frac{q}{L}\right) \\ & \Rightarrow \text { Work done }=\mathrm{Q}\left(\mathrm{~V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}\right) \\ & =-\frac{2}{3} \times \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}=\frac{-q Q}{6 \pi \varepsilon_0 L} \end{aligned}$

Hint

(c) Electric flux, $\phi_E=\int \vec{E} \cdot d \bar{S}$
$=\int E d S \cos \theta=\int E d S \cos 90^{\circ}=0 .$
The lines are parallel to the surface.

Hint

(b) Work done in deflecting a dipole through an angle $\theta$ is given by
$W=\int_0^\theta p E \sin \theta d \theta=p E(1-\cos \theta)$
Since $\theta=90^{\circ}$
$\therefore W=\mathrm{pE}\left(1-\cos 90^{\circ}\right) \Rightarrow W=\mathrm{pE}$

Hint

(a) Work done is equal to zero because the potential of A and B are the same $=\frac{1}{4 \pi \varepsilon_0} \frac{q}{a}$

No work is done if a particle does not change its potential energy. i.e. initial potential energy = final potential energy.

Hint

(c) Given the change in potential energy of the system is $\frac{q_3 k}{4 \pi \varepsilon_0}$. We have to find the value of $k$.
We know that potential energy of discrete system of charges is given by
$U=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{r_{12}}+\frac{q_2 q_3}{r_{23}}+\frac{q_3 q_1}{r_{31}}\right)$
According to question,
$\begin{aligned} & U_{\text {initial }}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{0.3}+\frac{q_2 q_3}{0.5}+\frac{q_3 q_1}{0.4}\right) \\ & U_{\text {final }}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{0.3}+\frac{q_2 q_3}{0.1}+\frac{q_3 q_1}{0.4}\right) \\ & U_{\text {final }}-U_{\text {initial }}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{0.1}-\frac{q_2 q_3}{0.5}\right) \\ & =\frac{1}{4 \pi \varepsilon_0}\left[10 q_2 q_3-2 q_2 q_3\right]=\frac{q_3}{4 \pi \varepsilon_0}\left(8 q_2\right) \dots(i) \end{aligned}$
$\text { But given, } \Delta \mathrm{U}=\frac{\mathrm{q}_3 \mathrm{k}}{4 \pi \varepsilon_{\mathrm{o}}} \dots(ii)$
$\begin{aligned} &\text { We get, }\\ &\mathrm{k}=\mathrm{q}_2(10-2)=8 \mathrm{q}_2 \end{aligned}$

Hint

(c) When the dipole is in the direction of field then net force is $q E+(-q E)=0$ and its potential energy is minimum $=-$ P.E. $=-\mathrm{qaE}$

Hint

(b)

$\begin{aligned} &\text { Using } \frac{1}{2} m v^2=q V\\ &V=\frac{1}{2} \times \frac{2 \times 10^{-3} \times 10 \times 10}{2 \times 10^{-6}}=50 \mathrm{kV} \end{aligned}$

Hint

(b) permittivity of free space $\epsilon_0$
$F=\frac{1 q_1 q_2}{4 \pi \epsilon_0 r^2}$; unit of $q$ is $C$, unit of $F$ is Newton and unit of $r$ is $m$
So the unit of $\epsilon_0$ is $\mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

Hint

(b) The total flux through the cube $\phi_{\text {total }}=\frac{q}{\varepsilon_0}$
$\therefore$ The electric flux through any face
$\phi_{\text {face }}=\frac{q}{6 \varepsilon_0}=\frac{4 \pi q}{6\left(4 \pi \varepsilon_0\right)}$

Hint

(c) Electric field intensity E is zero within a conductor due to charge given to it.
Also, $E=-\frac{d V}{d x} \Rightarrow \frac{d V}{d x}=0$ (inside the conductor)
$\therefore \mathrm{V}=$ constant. [V is potential]
So potential remains same throughout the conductor.

Hint

(c) There are eight corners of a cube and in each corner there is a charge of $(-\mathrm{q})$. At the centre of the corner there is a charge of $(+q)$. Each corner is equidistant from the centres of the cube and the distance (d) is half of the diagonals of the cube.
Diagonal of the cube $=\sqrt{b^2+b^2+b^2}=\sqrt{3} b$
$\therefore d=\sqrt{3} b / 2$
Now, electric potential energy of the charge $(+q)$ due to a charge $(-q)$ at one corner
$\mathrm{U}=\frac{q_1 q_2}{4 \pi \varepsilon_0 r}=\frac{(+q) \times(-q)}{4 \pi \varepsilon_0(\sqrt{3} b / 2)}=-\frac{q^2}{2 \pi \varepsilon_0(\sqrt{3} b)}$
$\therefore$ Total electric potential energy due to all the eight identical charges
$=8 U=-\frac{8 q^2}{2 \pi \varepsilon_0 \sqrt{3} b}=\frac{-4 q^2}{\sqrt{3} \pi \varepsilon_0 b}$

Hint

(a) For complete cube $\phi=\frac{Q}{\varepsilon_0} \times 10^{-6}$
For each face $\phi=\frac{1}{6} \frac{Q}{\varepsilon_0} \times 10^{-6}$

Hint

(b) Given : Dipole moment of the dipole $=\vec{p}$ and uniform electric field $=\vec{E}$. We know that dipole moment $(p)=\mathrm{q} . \mathrm{a}$ (where $q$ is the charge and $a$ is dipole length). And when a dipole of dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$, then Torque $(\tau)=$ Either force $\times$ Perpendicular distance between the two forces $=$ $q a E \sin \theta$ or $\tau=p E \sin \theta$ or $\vec{\tau}=\vec{p} \times \vec{E}$ (vector form)

Hint

(a) Given : Length of the dipole $(2 l)=10 \mathrm{~cm}$ $=0.1 \mathrm{~m}$ or $l=0.05 \mathrm{~m}$
Charge on the dipole $(\mathrm{q})=500 \mu \mathrm{C}=500 \times$ $10^{-6} \mathrm{C}$ and distance of the point on the axis from the mid-point of the dipole (r) $-20+5-25 \mathrm{~cm}-$ 0.25 m . We know that the electric field intensity due to dipole on the given point $(\mathrm{E})=$
$\begin{aligned} & \frac{1}{4 \pi \varepsilon_0} \times \frac{2(q .2 l) r}{\left(\mathrm{r}^2-l^2\right)^2} \\ & =9 \times 10^9 \times \frac{2\left(500 \times 10^{-6} \times 0.1\right) \times 0.25}{\left[(0.25)^2-(0.05)^2\right]^2} \\ & =\frac{225 \times 10^3}{3.6 \times 10^{-3}}=6.25 \times 10^7 \mathrm{~N} / \mathrm{C} \end{aligned}$
Note: The dipole field $E \propto \frac{1}{r^3}$ decreases much rapidly as compared to the field of a point charge $E \propto \frac{1}{r^2}$

Hint

(c) According to Gauss’s theorem, electric flux through a closed surface $=\frac{Q}{\varepsilon_0}$ where q is charge enclosed by the surface.

Hint

(c) $\lambda=latex] linear charge density; Charge on elementary portion [latex]d x=\lambda d x$.

Electric field at $O, d E=\frac{\lambda d x}{4 \pi \varepsilon_0 a^2}$
Horizontal electric field, i.e., perpendicular to $A O$, will be cancelled.
Hence, net electric field = addition of all electrical fields in direction of $A O$
$=\Sigma d E \cos \theta$
$\begin{aligned} & \Rightarrow E=\int \frac{\lambda d x}{4 \pi \varepsilon_0 a^2} \cos \theta \\ & \text { Also, } d \theta=\frac{d x}{a} \text { or } d x=a d \theta \\ & E=\int_{-\pi / 2}^{\pi / 2} \frac{\lambda \cos \theta d \theta}{4 \pi \varepsilon_0 a}=\frac{\lambda}{4 \pi \varepsilon_0 a}[\sin \theta]_{-\pi / 2}^{\pi / 2} \\ & =\frac{\lambda}{4 \pi \varepsilon_0 a}[1-(-1)]=\frac{\lambda}{2 \pi \varepsilon_0 a} \end{aligned}$

Hint

(b) Millikan evaluated the mass of electron indirectly with the help of charge by oil drop experiment.

Hint

(a)

In Milikan’s oil drop experiment, the charged drop falls with terminal velocity V . When an electric field E is applied in the vertically upward direction, the drop starts moving in the upward direction with terminal velocity 2 V . If the magnitude of the electric field is decreased to $\mathrm{E} / 2$, the terminal velocity will become V/2.

The terminal velocity is determined by balancing the gravitational force and the electric force acting on the drop. When the electric field is halved, the electric force is reduced by half, resulting in a decrease in terminal velocity to $\mathrm{V} / 2$.

Hint

(a) In air, $F_{\text {air }}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$ In medium, $F_m=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{K r^2}$
$\therefore \frac{F_m}{F_{\text {air }}}=\frac{1}{K} \Rightarrow F_m=\frac{F_{\text {air }}}{K}(\text { decreases K-times })$

Hint

$\begin{aligned} &\text { (d) Electric field at equatorial point }\\ &E e q=\frac{p}{4 \pi E_o\left(r^2+l^2\right)^{3 / 2}} \end{aligned}$
$\begin{aligned} &E=\frac{p}{4 \pi \varepsilon_0 \cdot r^3} \quad[\text { For } r \gg l]\\ &\text { Apparently, } E \propto p \text { and } E \propto \frac{1}{r^3} \propto r^{-3} \text {. } \end{aligned}$

Hint

(c) Let the initial dipole moment be $p$ and the initial distance be $r$. Thus, the initial electric field at the equator is:
$E=\frac{k \cdot p}{r^3}$
Doubling the Strength and Distance:
Now, if the dipole strength is doubled, it becomes $2 p$, and if the distance is also doubled, it becomes $2 r$.
Calculating the New Electric Field:
The new electric field $E^{\prime}$ at the equator with the new values will be:
$E^{\prime}=\frac{k \cdot(2 p)}{(2 r)^3}$
Simplifying the denominator:
$(2 r)^3=8 r^3$
Therefore, we can rewrite the new electric field as:
$E^{\prime}=\frac{k \cdot(2 p)}{8 r^3}$
Relating New Electric Field to Initial Electric Field:
Now, substituting the expression for $E$ :
$E^{\prime}=\frac{2 k \cdot p}{8 r^3}=\frac{1}{4} \cdot \frac{k \cdot p}{r^3}=\frac{1}{4} E$
Final Result: Thus, the new electric field $E^{\prime}$ when both the dipole strength and distance are doubled is:
$E^{\prime}=\frac{E}{4}$

Hint

(a) 1. Understand the Initial Conditions: The charge $q$ is initially at rest in the electric field. Therefore, the initial kinetic energy $K E_1=0$.
2. Apply the Work-Energy Theorem: The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy
$W=\Delta K E=K E_2-K E_1$
Since $K E_1=0$, we can simplify this to:
$W=K E_2$
3. Calculate the Work Done: The work done $W$ by the electric field on the charge as it moves through a distance $y$ can be calculated using the formula:
$W=F \cdot d$
Here, $F$ is the electric force acting on the charge, which can be expressed as:
$F=q E$
Therefore, the work done can be rewritten as:
$W=q E \cdot y$
4. Relate Work Done to Kinetic Energy: From the work-energy theorem, we have:
$K E_2=W=q E \cdot y$
5. Final Expression for Kinetic Energy: Thus, the kinetic energy of the charge after traveling a distance $y$ in the electric field $E$ is:
$K E=q E y$

Hint

(c) Electric field will be zero at the centre of hollow sphere. Charge resides on the outer surface of a conducting hollow sphere of radius R . We consider a spherical surface of radius $r<\mathrm{R}$. By Gauss’s theorem,

$\begin{aligned} &\begin{aligned} & \oint \vec{E} \cdot \vec{d} s=\frac{1}{\varepsilon_0} \times \text { charge enclosed } \\ & \text { or, } \quad E .4 \pi r^2=\frac{1}{\varepsilon_0} \times 0 \Rightarrow E=0 \end{aligned}\\ &\text { i.e. electric field inside a hollow sphere is zero. } \end{aligned}$
Note: Electric field inside the charged conducting sphere or shell, $E_{\text {in }}=0$
And for uniformly charged non-conducting sphere,
$E_{\text {in }}=\frac{1}{4 \pi \epsilon_0} \frac{Q_r}{R^3}$

Hint

(a) Electric field will be zero at the centre of hollow sphere. Charge resides on the outer surface of a conducting hollow sphere of radius R . We consider a spherical surface of radius $r<\mathrm{R}$. By Gauss’s theorem,

$\begin{aligned} &\begin{aligned} & \oint \vec{E} \cdot \vec{d} s=\frac{1}{\varepsilon_0} \times \text { charge enclosed } \\ & \text { or, } \quad E .4 \pi r^2=\frac{1}{\varepsilon_0} \times 0 \Rightarrow E=0 \end{aligned}\\ &\text { i.e. electric field inside a hollow sphere is zero. } \end{aligned}$
Note: Electric field inside the charged conducting sphere or shell, $E_{\text {in }}=0$
And for uniformly charged non-conducting sphere,
$E_{\text {in }}=\frac{1}{4 \pi \epsilon_0} \frac{Q_r}{R^3}$

Hint

(a) Given : Electric field (E) $=500 \mathrm{~V} / \mathrm{m}$ and potential difference $(\mathrm{V})=3000 \mathrm{~V}$. We know that electric field
$(\mathrm{E})=500=\frac{V}{d}$ or $d=\frac{3000}{500}=6 \mathrm{~m}$
[where $d=$ Distance between point charge and A]

Hint

$\begin{aligned} &\text { (c) Intensity of electric field due to a Dipole }\\ &\mathrm{E}=\frac{p}{4 \pi \varepsilon_0 r^3} \sqrt{3 \cos ^2 \theta+1} \Rightarrow E \propto \frac{1}{r^3} \end{aligned}$

Hint

(a) Dipole is formed when two equal and unlike charges are placed at a short distance.

Hint

$\begin{aligned} &\text { (a) By Gauss theorem }\\ &\begin{aligned} & \text { Total electric flux }=\frac{\text { Total charge inside cube }}{\varepsilon_0} \\ & \Rightarrow \phi=\frac{q}{\varepsilon_0} \end{aligned} \end{aligned}$

Hint

(c) Charges $(\mathrm{q})=2 \times 10^{-6} \mathrm{C}$,
Distance d $=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$ and
electric field (E)
$=2 \times 10^5 \mathrm{~N} / \mathrm{C}$
Torque $(\tau)=\mathrm{pE}=\mathrm{qdE}$
$\begin{aligned} & \mathrm{T}=\left(2 \times 10^{-6}\right) \times\left(3 \times 10^{-2}\right) \times\left(2 \times 10^5\right) \\ & =12 \times 10^{-3} \mathrm{~N}-\mathrm{m} . \end{aligned}$

Hint

(d) Charge $(\mathrm{q})=0.2 \mathrm{C}$;
Distance $\mathrm{d}=2 \mathrm{~m}$; Angle $\theta=60^{\circ}$ and work done
$(\mathrm{W})=4 \mathrm{~J}$
Work done in moving the charge (W)
$=F \cdot d \cos \theta=q E d \cos \theta$
$\text { or, } \begin{aligned} E & =\frac{W}{q d \cos \theta}=\frac{4}{0.2 \times 2 \times \cos 60^{\circ}}=\frac{4}{0.4 \times 0.5} \\ & =20 \mathrm{~N} / \mathrm{C} \end{aligned}$

Hint

(c) Net force on each of the charge due to the other charges is zero. However, disturbance in any direction other than along the line on which the charges lie, will not make the charges return.

Hint

(d) Charges (-e) on electron and (e) on proton exert a force of attraction given by
$\text { Force }=(K) \frac{(-e)(e)}{r^2} \hat{r}=\frac{-K e^2}{r^3} \vec{r} \quad\left(\because \hat{r}=\frac{\vec{r}}{|r|}\right)$

Note: Magnitude of Coulomb force is given by $\frac{1}{4 \pi \varepsilon_0} \frac{q_2 q_2}{r^2}$ but in vector form $\vec{F}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^3} \vec{r}$