Entrance Corner

Hint

(b) Given an electric flux \(\phi=-2 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1}\) through a spherical Gaussian surface with a radius of \(8.0 \mathrm{~cm}(r=0.08 \mathrm{~m})\), we are to find the value of the point charge \(Q\).
According to Gauss’s Law: \(\phi=\frac{Q}{\varepsilon_0}\)
Where \(\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\).
Rearranging the equation to solve for \(Q\), we have:
\(
Q=\phi \varepsilon_0
\)
Substituting the given values:
\(
Q=-2 \times 10^4 \times 8.85 \times 10^{-12}
\)
Calculating this yields:
\(
Q=-17.7 \times 10^{-8} \mathrm{C}
\)
Thus, the value of the point charge is \(Q=-17.7 \times 10^{-8} \mathrm{C}\).

Hint

(c) We know, electric field due to uniformly charged infinite plane sheet,
\(
E=\frac{\sigma_0}{2 \varepsilon_0}
\)
\(
\vec{\tau}=\vec{{P}} \times \vec{{E}}
\)
The electric field is uniform, and there is no variation in the field across the dipole, meaning the torque on the dipole is zero.
Also, the dipole is in the minimum potential energy configuration when aligned with the electric field, and the net force on the dipole due to the uniform electric field is zero.
Therefore, the potential energy of the dipole is minimum, and the torque is zero.

Hint

(b) The torque acting on a dipole in a uniform electric field is:
\(
\begin{gathered}
\tau=p \times E \\
\tau=p E_0 \sin \theta
\end{gathered}
\)
\(
\text { Dipole moment: } p=q l
\)
\(
\tau=q l E_0 \sin \theta
\)
For small angles, \(\sin \theta \approx \theta\), so:
\(
\tau \approx q l E_0 \theta
\)
Using Newton’s second law for rotational motion:
\(
I \alpha=\tau
\)
\(\alpha=\frac{d^2 \theta}{d t^2}\) is the angular acceleration.
For a dipole (two point masses \(q\) separated by \(l\) ), the moment of inertia about the center is:
\(
I=m\left(\frac{l}{2}\right)^2+m\left(\frac{l}{2}\right)^2=2 m \frac{l^2}{4}=\frac{m l^2}{2}
\)
So the equation of motion becomes:
\(
\begin{aligned}
& \frac{m l^2}{2} \frac{d^2 \theta}{d t^2}=-q l E_0 \theta \\
& \frac{d^2 \theta}{d t^2}+\frac{2 q E_0}{m l} \theta=0
\end{aligned}
\)
This is the standard form of simple harmonic motion:
\(
\frac{d^2 \theta}{d t^2}+\omega^2 \theta=0
\)
where:
\(
\omega^2=\frac{2 q E_0}{m l}
\)
Thus, the angular frequency is:
\(
\omega=\sqrt{\frac{2 q E_0}{m l}}
\)
The time period of oscillation is given by:
\(
\begin{gathered}
T=\frac{2 \pi}{\omega} \\
T=2 \pi \sqrt{\frac{m l}{2 q E_0}}
\end{gathered}
\)

Hint

(d)

(A) \(\rightarrow 0\) (III)
(B) \(\rightarrow \frac{\sigma}{2 \varepsilon_0}\)
(C) \(\rightarrow \frac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{I}^2}\) (No row matching)
(D) \(\rightarrow \frac{\sigma}{\varepsilon_0}({I})\)

Hint

(a) Electric field due to infinitely long wire
\(
\vec{E}=-\frac{2 k \lambda}{r} \hat{r}
\)
We know, \(V=-\int \vec{E} \cdot \overrightarrow{d r}\)
\(
\begin{aligned}
& =-\int \frac{-2 k \lambda}{r} \hat{r} \cdot \overrightarrow{d r} \\
& V=\int \frac{2 k \lambda}{r} d r=2 k \lambda \ln r+c
\end{aligned}
\)
Net potential due to all wires,
\(
\begin{aligned}
& V=v_1+v_2+v_3 \\
& =2 k \lambda \ln \left(\sqrt{x^2+y^2}\right)+2 k \lambda \ln \left(\sqrt{y^2+z^2}\right)+2 k \lambda \ln \left(\sqrt{z^2+x^2}\right)+c
\end{aligned}
\)
for equipotential surface, \(V=c\)
\(
\begin{aligned}
& \Rightarrow 2 k \lambda\left[\ln \left(\sqrt{x^2+y^2} \cdot \sqrt{y^2+z^2} \cdot \sqrt{z^2+x^2}\right)\right]=c \\
& \Rightarrow \sqrt{\left(x^2+y^2\right) \cdot\left(y^2+z^2\right) \cdot\left(z^2+x^2\right)}=c \\
& \Rightarrow\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=c
\end{aligned}
\)
where, c = constant.

Hint

(c)

Explanation:
To find the speed of the particle when it crosses the \(x\)-axis, we can use the concept of energy conservation. The work done by the electric field on the particle will be equal to the change in kinetic energy of the particle.
The force acting on the particle due to the electric field is given by \(F=q E\). The work done by this force when the particle moves a distance \(L\) is \(W=F \cdot d=q E \cdot L\).
The change in kinetic energy is given by \(\Delta K E=\frac{1}{2} m v^2-O=\frac{1}{2} m v^2\). Setting the work done equal to the change in kinetic energy gives us: \(q E \cdot L=\frac{1}{2} m v^2\).
Rearranging this equation to solve for \(v\) gives us: \(v=\sqrt{\frac{2 q E L}{m}}\). Therefore, the speed of the particle when it crosses the x-axis is \(v=\sqrt{\frac{2 q E L}{m}}\).

Step by Step Solution:
Step 1: Identify the forces acting on the particle. The only force acting on the particle is the electric force due to the electric field. which is \(F=q E\).

Step 2: Calculate the work done by the electric field when the particle moves a distance \(L\) : \(W=F \cdot d=q E \cdot L\).

Step 3: Use the work-energy principle: the work done is equal to the change in kinetic energy. Set the work done equal to the change in kinetic energy: \(q E \cdot L=\frac{1}{2} m v^2\).

Step 4: Rearrange the equation to solve for the speed \(v: v=\sqrt{\frac{2 q E L}{m}}\).
Final Answer: \(v=\sqrt{\frac{2 q E L}{m}}\)

Hint

(b) Charge on one sphere: \(q_1=4 \times 10^{-8} \mathrm{C}\)
Force of repulsion: \(F=9 \times 10^{-3} \mathrm{~N}\)
Constant: \(k=\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^9 \frac{\mathrm{~N} \cdot \mathrm{~m}^2}{\mathrm{C}^2}\)
When two identical conducting spheres come into contact, the charge is evenly distributed between them.
Coulomb’s law: \(F=k \frac{\left|q_1 q_2\right|}{r^2}\)
Calculate the charge on each sphere after contact

When the uncharged sphere touches the charged sphere, the charge is distributed equally.

Each sphere will have a charge of:
\(q=\frac{q_1+0}{2}=\frac{4 \times 10^{-8} \mathrm{C}}{2}=2 \times 10^{-8} \mathrm{C}\)
Apply Coulomb’s law
The force between the spheres is given by:
\(F=k \frac{q^2}{r^2}\)

Where \(r\) is the distance between the spheres.
Solve for the distance \(r\)

Rearrange the formula to solve for \(r\) :
\(r^2=k \frac{q^2}{F}\)
\(r=\sqrt{k \frac{q^2}{F}}\)
\(
\begin{aligned}
&\text { Substitute the values and calculate } r\\
&\begin{aligned}
r & =\sqrt{\left(9 \times 10^9 \frac{\mathrm{~N} \cdot \mathrm{~m}^2}{\mathrm{C}^2}\right) \frac{\left(2 \times 10^{-8} \mathrm{C}\right)^2}{9 \times 10^{-3} \mathrm{~N}}} \\
r & =\sqrt{\left(9 \times 10^9\right) \frac{4 \times 10^{-16}}{9 \times 10^{-3}} \mathrm{~m}} \\
r & =\sqrt{4 \times 10^{-4}} \mathrm{~m} \\
r & =2 \times 10^{-2} \mathrm{~m} \\
r & =2 \mathrm{~cm}
\end{aligned}
\end{aligned}
\)

Hint

(d)

\(
\begin{gathered}
\mathrm{U}_1=\frac{4 \mathrm{Kq}_0^2}{\mathrm{a}}+\frac{2 \mathrm{Kq}_0^2}{\sqrt{2} \mathrm{a}}=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4+\sqrt{2}) \\
\mathrm{U}_2=\frac{\mathrm{Kq}_0^2}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}(4+\sqrt{2})=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4 \sqrt{2}+2) \\
\mathrm{U}_2-\mathrm{U}_1=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)
\end{gathered}
\)

Hint

(d)

\(
u=\frac{1}{2} \varepsilon_0 E^2
\)
The energy density of an electric field in free space is given by the expression above. For a parallel plate capacitor with plate area \(A\) and separation \(d\), the volume between the plates is
\(
V=A d
\)
Multiplying the energy density by the volume gives the total energy stored:
\(
U=u \times V=\frac{1}{2} \varepsilon_0 E^2 \times A d=\frac{1}{2} \varepsilon_0 E^2 A d
\)
Thus, the potential energy stored in the capacitor is
\(
\frac{1}{2} \varepsilon_0 E^2 A d
\)

Hint

(d) The work done on an electric dipole in a uniform electric field when it is rotated from an angle \(\theta_1\) to an angle \(\theta_2\) is given by:
\(
W=-p E\left(\cos \theta_2-\cos \theta_1\right)
\)
where:
\(p=q \cdot d\) is the dipole moment,
\(E\) is the electric field strength,
\(\theta_1\) and \(\theta_2\) are the initial and final angles between the dipole moment and the electric field.
Given:
Charge \(q=4 \mu \mathrm{C}=4 \times 10^{-6} \mathrm{C}\)
Distance \(d=18 \mathrm{~cm}=0.18 \mathrm{~m}\)
Electric field \(E=10^4 \mathrm{~N} / \mathrm{C}\)
Initial angle \(\theta_1=0^{\circ}\) (aligned with the field, equilibrium position)
Final angle \(\theta_2=180^{\circ}\)
First, calculate the dipole moment \(p\) :
\(
p=q \cdot d=4 \times 10^{-6} \mathrm{C} \cdot 0.18 \mathrm{~m}=7.2 \times 10^{-7} \mathrm{Cm}
\)
Calculate the work done \(W\) :
\(
\begin{aligned}
& W=-p E\left(\cos 180^{\circ}-\cos 0^{\circ}\right) \\
& W=-7.2 \times 10^{-7} \mathrm{Cm} \cdot 10^4 \mathrm{~N} / \mathrm{C} \cdot(-1-1) \\
& W=7.2 \times 10^{-7} \mathrm{Cm} \cdot 10^4 \mathrm{~N} / \mathrm{C} \cdot 2 \\
& W=1.44 \times 10^{-2} \mathrm{~J}=14.4 \mathrm{~mJ}
\end{aligned}
\)

Hint

(d) Charge \(q_1=7 \mu \mathrm{C}=7 \times 10^{-6} \mathrm{C}\) at \((-7 \mathrm{~cm}, 0,0)\).
Charge \(q_2=-4 \mu \mathrm{C}=-4 \times 10^{-6} \mathrm{C}\) at \((7 \mathrm{~cm}, 0,0)\).
Permittivity of free space \(\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\).
The electrostatic potential energy \(U\) between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by \(U=\frac{1}{4 \pi e_0} \frac{q_1 q_2}{r}\).
Step 1: Calculate the distance \(r\) between the charges.
The coordinates of the charges are \((-0.07 \mathrm{~m}, 0,0)\) and \((0.07 \mathrm{~m}, 0,0)\).
The distance between them is \(r=0.07 \mathrm{~m}-(-0.07 \mathrm{~m})=0.14 \mathrm{~m}\).

Step 2: Calculate the electrostatic potential energy \(U\).
\(
\begin{aligned}
& U=\frac{1}{4 \pi e_0} \frac{q_1 q_2}{r} \\
& U=\frac{1}{4 \pi\left(8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \frac{\left(7 \times 10^{-6} \mathrm{C}\right)\left(-4 \times 10^{-6} \mathrm{C}\right)}{0.14 \mathrm{~m}} \\
& U=\left(8.99 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \frac{-28 \times 10^{-12} \mathrm{C}^2}{0.14 \mathrm{~m}} \\
& U=\frac{-0.25172}{0.14} \mathrm{~J} \\
& U=-1.798 \mathrm{~J} \\
& U \approx-1.8 \mathrm{~J}
\end{aligned}
\)
The electrostatic potential energy of the charge configuration is approximately -1.8 J.

Hint

(b)

\(
\begin{aligned}
& \frac{\mathrm{kq}}{(\mathrm{r}-\mathrm{a})^2}=\frac{\mathrm{kq}}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{kq}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)} \cos \theta \\
& \frac{1}{(\mathrm{r}-\mathrm{a})^2}=\frac{1}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}} \\
& \frac{1}{(\mathrm{r}-\mathrm{a})^2}-\frac{1}{(\mathrm{r}+\mathrm{a})^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}} \\
& \frac{4 \mathrm{ra}}{\left(\mathrm{r}^2-\mathrm{a}^2\right)^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}} \\
& \Rightarrow \frac{2 r}{\left(r^2-a^2\right)^2}=\frac{1}{\left(r^2+a^2\right)^{\frac{3}{2}}} \\
& \frac{4 r^2}{\left(r^2-a^2\right)^4}=\frac{1}{\left(r^2+a^2\right)^3} \\
& \Rightarrow 4 r^2\left(r^2+a^2\right)^3=\left(r^2-a^2\right)^4 \\
& 4 r^8\left(1+\frac{a^2}{r^2}\right)^3=r^8\left(1-\frac{a^2}{r^2}\right)^4 \\
& 4\left(1+\frac{a^2}{r^2}\right)^3=\left(1-\frac{a^2}{r^2}\right)^4
\end{aligned}
\)
\(
\text { But by solving from mathematical software we are getting } \mathrm{a} / \mathrm{r} \approx 3 \text {. }
\)

Hint

(a) We are given that the electric flux is:
\(
\varphi=\alpha \sigma+\beta \lambda
\)
Here, \(\alpha\) and \(\beta\) are constants, \(\sigma\) is the surface charge density, and \(\lambda\) is the linear charge density. Now, let’s take the dimensions of both sides of the equation. For electric flux \(\varphi\), the dimension is:
\(
\begin{gathered}
{[\varphi]=[\alpha \sigma]=[\beta \lambda]} \\
{[\alpha]=\left[\frac{\varphi}{\sigma}\right]=\left[\frac{[Q / L]}{[Q / \text { Area }]}\right]=\left[\frac{\text { Area }}{\text { Length }}\right]}
\end{gathered}
\)
So, the dimensions of \(\alpha\) are \(\left[\frac{L^2}{L}\right]=L\). For \(\beta\), we get:
\(
[\beta]=\left[\frac{\varphi}{\lambda}\right]=\left[\frac{[Q / L]}{[Q / \text { Area }]}\right]=[L]
\)
Thus, the quantity \(\frac{\alpha}{\beta}\) represents a length, which corresponds to a displacement. Therefore, the correct answer is (a) displacement.

Hint

(c) Electric Potential at B:
For a short dipole, the electric potential at points on the equatorial plane (like point B) is zero.

Electric Field at B:
The electric field due to a dipole at a point on the x -axis (like point A ) is given by \(E=(2 \mathrm{kp}) / \mathrm{r}^3\), where k is Coulomb’s constant, p is the dipole moment, and r is the distance.
Similarly, the electric field at a point on the \(y\)-axis (like point \(B\) ) is given by \(E=\) \((k p) / r^3\).
Since the distance from the dipole to point B is 4 times the distance from the dipole to point \(A(r B=4 r A)\), the electric field at \(B\) is \(1 / 16\) th of the electric field at \(A\) ( \(E_B=(1 / 16) E_0\) ).

Conclusion: Therefore, the electric potential at point B is 0 , and the electric field at point \(B\) is \(\mathrm{E}_0 / 16\).

Hint

(b) We can solve this using Gauss’s law, which states that the total electric flux \(\Phi\) through a closed surface is proportional to the total charge enclosed within the surface.
Mathematically:
\(
\Phi=\frac{Q_{\text {enclosed }}}{\epsilon_0}
\)
Step 1: Determine the enclosed charge \(Q_{\text {enclosed }}\)
Assume three imaginary cubes (total four cubes) surrounding the edge BC. The given line charge has a linear charge density \(\lambda\) (charge per unit length) and length \(\frac{a}{2}\). Therefore, the total charge \(Q_{\text {enclosed }}\) on the line is:
\(
Q_{\text {enclosed }}=\lambda \cdot \frac{a}{2}
\)
Step 2: Apply Gauss’s law
The total electric flux through all the four cubes is:
\(
\begin{gathered}
\Phi=\frac{Q_{\text {enclosed }}}{\epsilon_0} \\
\Phi=\frac{\lambda \cdot \frac{a}{2}}{\epsilon_0} \\
\Phi=\frac{\lambda a}{2 \epsilon_0}
\end{gathered}
\)
Final Answer: The total electric flux through one cube is:
\(
\Phi_{\text {single cube }}=\frac{\Phi}{4}=\frac{\frac{\lambda a}{2 \epsilon_0}}{4}=\frac{\lambda a}{8 \epsilon_0}
\)

Hint

(a) Potential energy of a dipole in an electric field: \(U=-p E \cos \theta\)
Work done is equal to the change in potential energy: \(W=\Delta U=U_2-U_1\)
\(
\begin{aligned}
& U_1=-p E \cos \theta_1 \\
& U_1=-\left(6 \times 10^{-6} \mathrm{Cm}\right)\left(10^6 \mathrm{~V} / \mathrm{m}\right) \cos \left(0^{\circ}\right) \\
& U_1=-\left(6 \times 10^{-6}\right)\left(10^6\right)(1) \\
& U_1=-6 \mathrm{~J}
\end{aligned}
\)
\(
\begin{aligned}
& U_2=-p E \cos \theta_2 \\
& U_2=-\left(6 \times 10^{-6} \mathrm{Cm}\right)\left(10^6 \mathrm{~V} / \mathrm{m}\right) \cos \left(180^{\circ}\right) \\
& U_2=-\left(6 \times 10^{-6}\right)\left(10^6\right)(-1) \\
& U_2=6 \mathrm{~J}
\end{aligned}
\)
\(
\begin{aligned}
& W=\Delta U=U_2-U_1 \\
& W=6 \mathrm{~J}-(-6 \mathrm{~J}) \\
& W=12 \mathrm{~J}
\end{aligned}
\)

Hint

(c)

Total flux through square \(=\frac{\mathrm{q}}{\epsilon_0}\left(\frac{1}{6}\right)\)
Lets divide square is 8 equal parts.
Flux is same for each part.
\(\therefore\) Flux through shaded portion is \(\frac{8}{5}\) (Total flux)
\(
=\frac{5}{8} \times \frac{q}{\epsilon_0} \frac{1}{6}=\frac{5}{48} \frac{1}{\epsilon_0}
\)
\(\therefore\) answer is 48

Hint

(b) To find the ratio of the magnitudes of electrostatic force to gravitational force between ions \(A\) and \(B\), we use the following formulas for electrostatic force \(\left(F_e\right)\) and gravitational force \(\left(F_g\right)\) :
\(
\begin{aligned}
& F_e=\frac{k \cdot q_1 \cdot q_2}{r^2} \\
& F_g=\frac{G \cdot m_1 \cdot m_2}{r^2}
\end{aligned}
\)
To find the ratio \(\frac{F_e}{F_g}\), simplify as follows:
\(
\frac{F_e}{F_g}=\frac{k \cdot q_1 \cdot q_2}{G \cdot m_1 \cdot m_2}
\)
\(
\frac{F_e}{F_g}=\frac{9 \times 10^9 \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}}
\)
\(
=\frac{10^{-20}}{2 \times 10^{-65}}=5 \times 10^{44}
\)
This calculation gives the ratio of the electrostatic force to the gravitational force without considering their separation distance \(r\), as it cancels out. The value of \(P[latex] mentioned in the prompt can be deduced from the simplified final expression. [latex]\text { Thus, } P=10 \text {. }\)

Hint

(d) According to Gauss’s law, the electric flux \(\Phi\) through a closed surface is given by:
\(
\Phi=\frac{q_{\text {in }}}{\epsilon_0}
\)
where \(q_{\text {in }}\) is the total charge enclosed by the surface.
In this problem, the charges enclosed by the surface are \(q_{i n}=q+(-2 q)+5 q=4 q\). Thus, the electric flux is:
\(
\Phi=\frac{4 q}{\epsilon_0}
\)

Hint

(b) Let a be the radius of a sphere \(\mathrm{A}, Q_A\) be the charge on the sphere, and \(C_A\) be the capacitance of the sphere. Let b be the radius of a sphere \(\mathrm{B}, Q_B\) be the charge on the sphere, and \(C_B\) be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential \((V)\) will become equal.
Let \(E_A\) be the electric field of sphere A and \(E_B\) be the electric field of sphere B. Therefore, their ratio,
\(
\begin{aligned}
& \frac{E_A}{E_B}=\frac{Q_A}{4 \pi \epsilon_0 \times a_2} \times \frac{b^2 \times 4 \pi \epsilon_0}{Q_B} \\
& \frac{E_A}{E_B}=\frac{Q_A}{Q_B} \times \frac{b^2}{a^2} \ldots \text { (1) }
\end{aligned}
\)
However, \(\frac{Q_A}{Q_B}=\frac{C_A V}{C_B V}\)
And, \(\frac{C_A}{C_B}=\frac{a}{b}\)
\(
\therefore \frac{Q_A}{Q_B}=\frac{a}{b} \ldots(2)
\)
Putting the value of (2) in (1), we obtain
\(
\therefore \frac{E_A}{E_B}-\frac{a}{B} \frac{b^2}{a^2}=\frac{b}{a}
\)
Therefore, the ratio of electric fields at the surface is \(\frac{b}{a}\).

Hint

(b) Step 1: Initially, the force of repulsion between P and S is 16 N .
Step 2: Sphere R is brought into contact with P , and the charge is shared equally. Charge on P and R after contact: \(\frac{Q}{2}\).
Step 3: Sphere R with charge \(\frac{Q}{2}\) is brought into contact with S . Charge on S and R after contact: \(\frac{3 Q}{4}\).
Step 4: After contact, the charges on P and S are \(\frac{Q}{2}\) and \(\frac{3 Q}{4}\) respectively.
Step 5: Calculate the new force of repulsion using Coulomb’s law: \(F^{\prime}=k \frac{\left(\frac{q}{2}\right)\left(\frac{3 q}{2}\right)}{r^2}=\frac{3}{8} F=6 \mathrm{~N}\).

Hint

(b) Gauss’s law: \(\oint \vec{E} \cdot d \vec{A}=\frac{Q_{e n c}}{\epsilon_0}\)
Surface area of a sphere: \(A=4 \pi R^2\)
Consider a spherical Gaussian surface of radius \(r(r \approx R)\) just outside the charged shell.

The electric field \(\vec{E}\) is radial and constant in magnitude over the Gaussian surface.
The enclosed charge is \(Q_{\text {enc }}=\sigma A=\sigma\left(4 \pi R^2\right)\).
Applying Gauss’s law:
\(\oint \vec{E} \cdot d \vec{A}=E \oint d A=E\left(4 \pi R^2\right)=\frac{Q_{\text {enc }}}{\epsilon_0}\)
\(E\left(4 \pi R^2\right)=\frac{\sigma\left(4 \pi R^2\right)}{\epsilon_0}\)
Divide both sides by \(4 \pi R^2\) :
\(E=\frac{\sigma}{\epsilon_0}\)
The electric field at any point on the surface of the spherical shell is \(\frac{\sigma}{\epsilon_0}\).

Hint

(d) Option 1: Control the speed of the vehicle. This option is incorrect. Metallic ropes touching the ground during motion do not control the speed of the vehicle.

Option 2: Conduct the charge produced by friction. This is the correct answer. When vehicles carrying inflammable materials move, friction between the vehicle and the ground can generate static electricity. This static electricity can potentially ignite the inflammable materials, leading to a fire or explosion. By having metallic ropes touching the ground, the charge produced by friction is conducted away, preventing a buildup of static electricity and reducing the risk of ignition.

Option 3: Conduct the current produced by inflammable material. This option is incorrect. Inflammable materials do not typically produce current. The main concern is the generation of static electricity.

Option 4: Provide earthing for lightning. This option is incorrect. Metallic ropes touching the ground during motion do not provide earthing for lightning. Lightning protection systems use different mechanisms to safely direct lightning strikes to the ground.

In conclusion, option 2 is the correct answer as metallic ropes touching the ground during motion help conduct the charge produced by friction, reducing the risk of ignition for vehicles carrying inflammable materials.

Hint

(b)

\(
\begin{gathered}
\mathrm{F}_{\mathrm{e}}=\frac{\mathrm{KQ}_1 Q_2}{\mathrm{r}_2} \& \mathrm{~F}_{\mathrm{g}}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{\mathrm{r}_2} \\
\frac{\mathrm{~F}_{\mathrm{e}}}{\mathrm{~F}_{\mathrm{g}}}=\frac{9 \times 10^9 \times 1.6 \times 10^{-19^2}}{6.67 \times 10^{-11} 9.1 \times 10^{-31} 1.67 \times 15^{-27}}=2.27 \times 10^{39}
\end{gathered}
\)

Hint

(a) Gauss’s law states that the total electric flux through a closed surface is directly proportional to the enclosed electric charge: \(\Phi=\frac{q_{\text {enclosed }}}{\epsilon_0}\).
To apply Gauss’s law effectively, imagine a closed surface enclosing the charge.
Imagine another identical cube placed adjacent to the first cube, with the charge \(q\) at their common face.
This creates a closed Gaussian surface (a cube of double the volume) enclosing the charge \(q\).

The total flux through the entire doubled cube is given by Gauss’s law:
\(
\Phi_{\text {total }}=\frac{q}{e_0}
\)
Since the original cube is half of the doubled cube, the flux through it is half of the total flux:
\(
\begin{aligned}
\Phi_{\text {cube }} & =\frac{1}{2} \Phi_{\text {total }} \\
\Phi_{\text {cube }} & =\frac{1}{2} \frac{q}{\varepsilon_0} \\
\Phi_{\text {cube }} & =\frac{q}{2 \varepsilon_0}
\end{aligned}
\)
The electric flux linked with the cube is \(\frac{q}{2 \epsilon_0}\).

Hint

(c) Electric field E at a distance r due to infinite long wire is \(E=\frac{2 k \lambda}{r}\)
Force of electron \(\Rightarrow F=e E\)
\(
\begin{aligned}
& \mathrm{F}=\mathrm{e}\left(\frac{2 \mathrm{k} \lambda}{\mathrm{r}}\right) \\
& \mathrm{F}=\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{r}}
\end{aligned}
\)
This force will provide required centripetal force
\(
\begin{aligned}
& \mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{r}} \\
& \mathrm{v}=\sqrt{\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{~m}}} \\
& \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m}\left(\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{~m}}\right) \\
& =\mathrm{k} \lambda \mathrm{e}
\end{aligned}
\)

Hint

(c) The electric flux \((\Phi)\) through a closed surface, according to Gauss’s law, is proportional to the enclosed charge \(Q\), and is given by \(\Phi=\frac{Q}{\epsilon_0}\), where \(\epsilon_0\) is the vacuum permittivity. This is regardless of the shape of the surface, as long as it is closed and it encloses the charge fully.
In this case, both cubes \(C_1\) and \(C_2\) are closed surfaces, and they are concentric, meaning they share the same center. Each cube encloses a different charge, \(2 Q\) for \(C_1\) and \(3 Q+\) \(2 Q=5 Q\) for \(C_2\). According to Gauss’s law, the electric flux through each cube is directly proportional to the enclosed charge.
For \(C_1\), enclosing charge \(2 Q: \Phi_{C_1}=\frac{2 Q}{\epsilon_0}\)
For \(C_2\), enclosing charge \(5 Q: \Phi_{C_2}=\frac{5 Q}{\epsilon_0}\)
Now, we want the ratio of electric flux through \(C_1\) to that through \(C_2\) :
\(
\frac{\Phi_{C_1}}{\Phi_{C_2}}=\frac{\frac{2 Q}{\epsilon_0}}{\frac{5 Q}{\epsilon_0}}=\frac{2 Q}{5 Q}=\frac{2}{5}
\)
Therefore, the ratio of electric flux passing through \(C_1\) to that passing through \(C_2\) is \(\frac{2}{5}\), which corresponds to option C.

Hint

(a) To find the force between two point charges in a medium with dielectric constant \(K\), we use Coulomb’s Law modified for the medium. The force in a medium is given by:
\(
F_m=\frac{F}{K}
\)
where \(F\) is the force in vacuum. Additionally, if the distance between the charges changes, the force is inversely proportional to the square of the distance. Therefore, the new force can be calculated as:
\(
F^{\prime}=\frac{F_m}{\left(\frac{r}{r^{\prime}}\right)^2}
\)
where \(r^{\prime}\) is the new distance.
Step 1: The initial force between the charges in vacuum is given by Coulomb’s Law:
\(
F=\frac{k q_1 q_2}{r^2}
\)
Step 2: When placed in a medium with dielectric constant \(K=5\), the force is reduced by a factor of \(K\) :
\(
F_m=\frac{F}{K}=\frac{F}{5}
\)
Step 3: The new distance between the charges is \(\frac{r}{5}\). The force is inversely proportional to the square of the distance:
\(
F^{\prime}=F_m\left(\frac{r}{r^{\prime}}\right)^2=\frac{F}{5}\left(\frac{r}{\frac{r}{5}}\right)^2=\frac{F}{5} \cdot 25=5 F
\)
Step 4: Thus, the force between the charges in the medium at the new distance is:
\(
F^{\prime}=5 F
\)

Hint

(c)

The electric field due to charge \(q\) at distance \(x\) is \(E_1=k \frac{|q|}{x^2}\).
The electric field due to charge \(3 q\) at distance \(r-x\) is \(E_2=k \frac{|3 q|}{(r-x)^2}\).
For the resultant electric field to be zero, the magnitudes of \(E_1\) and \(E_2\) must be equal:
\(E_1=E_2\)
\(k \frac{|q|}{x^2}=k \frac{|3 q|}{(r-x)^2}\)
Cancel out \(k\) and \(|q|\) from both sides:
\(\frac{1}{x^2}=\frac{3}{(r-x)^2}\)

Take the square root of both sides:
\(\frac{1}{x}=\frac{\sqrt{3}}{r-x}\)

Solve for \(x\) :
\(r-x=\sqrt{3} x\)
\(r=x+\sqrt{3} x\)
\(r=x(1+\sqrt{3})\)
\(x=\frac{r}{1+\sqrt{3}}\)
The distance \(x\) from charge \(q\) where the resultant electric field is zero is \(\frac{r}{1+\sqrt{3}}\).

Hint

(d) Electric field due to an infinitely long line charge: \(E=\frac{2 k \lambda}{r}\)
Centripetal force: \(F_c=\frac{m v^2}{r}\)
Relationship between velocity and time period: \(v=\frac{2 \pi r}{T}\)
The electric field due to the line charge is \(E=\frac{2 k \lambda}{r}\).
The electrostatic force on the particle is \(F_e=q E=\frac{2 k \lambda q}{r}\).
The centripetal force is \(F_c=\frac{m v^2}{r}\).
Equating the forces: \(\frac{2 k \lambda q}{r}=\frac{m v^2}{r}\).
\(
\begin{aligned}
& v^2=\frac{2 k \lambda q}{m} \\
& v=\sqrt{\frac{2 k \lambda q}{m}}
\end{aligned}
\)
The relationship between velocity and time period is \(v=\frac{2 \pi r}{T}\).
Substituting the value of \(v: \sqrt{\frac{2 k \lambda q}{m}}=\frac{2 \pi r}{T}\).
Solving for \(T: T=\frac{2 \pi r}{\sqrt{\frac{2 k \lambda q}{m}}}=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}\).

Hint

(c) The electric potential \(\boldsymbol{v}\) due to a dipole at a distance \(r\) is given by:
\(V=\frac{1}{4 \pi \epsilon_0} \frac{p \cos \theta}{r^2}\)
where \(p\) is the dipole moment, \(\theta\) is the angle between the dipole moment and the position vector, and \(\epsilon_0\) is the permittivity of free space.
From the formula, it’s clear that the potential \(V\) is proportional to \(\frac{1}{r^2}\).
\(V \propto \frac{1}{r^2}\)

Hint

(c)

\(
\begin{aligned}
& \overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{~A}}=30 \hat{\mathrm{i}} \\
& \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~A}} \\
& \phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mathrm{i}}) \\
& \phi=6 \times 30=180
\end{aligned}
\)

Hint

(d) Determine which charges are inside the sphere.
The sphere radius is \(4 a\).
The charge \(5 Q\) is at \((3 a, 0)\), which is inside the sphere since \(3 a<4 a\).
The charge \(-2 Q\) is at \((-5 a, 0)\), which is outside the sphere since \(5 a>4 a\).

Calculate the total charge enclosed by the sphere.
\(Q_{\text {enclosed }}=5 Q\)
Use Gauss’s law to find the electric flux.
\(\phi=\frac{Q_{\text {enclosed }}}{\epsilon_0}\)
\(\phi=\frac{5 Q}{\epsilon_0}\)

The electric flux through the sphere is \(\frac{5 Q}{\epsilon_0}\).

Hint

(a) The work done by an electric field when moving a charge on an equipotential surface is zero because there is no change in electric potential energy. Electric lines of force are always perpendicular to equipotential surfaces, meaning that any movement along the surface does not alter the electric potential, confirming that no work is done.

\(
W=q \Delta V \cos \theta
\)
Because \(\Delta V=0\) on an equipotential surface, irrespective of the value of \(\cos \theta\), the work \(W\) will be zero. Hence, the Assertion \((\mathrm{A})\) is correct.

Hint

(a)
\(
\begin{aligned}
& \text { Potential difference }=\frac{K Q}{r_1}-\frac{K Q}{r_2} \\
& r_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2} \\
& r_2=\sqrt{(\sqrt{6})^2+0}
\end{aligned}
\)
As \(r_1=r_2=\sqrt{6} \mathrm{~m}\)
So potential difference \(=0\)

Hint

(c) The electric flux is given by Gauss’s Law:
\(
\Phi=\oint \vec{E} \cdot d \vec{A}
\)
The field \(\vec{E}=2 x \hat{i}\) varies with \(x\).
For the cube, only the left ( \(x=0\) ) and right ( \(x=2\) ) faces contribute: \(\Phi=E_{\text {right }} A-E_{\text {left }} A\).
Substituting \(A=4 \mathrm{~m}^2, E_{\text {right }}=2(2)=4\), and \(E_{\text {left }}=2(0)=0\) :
\(
\Phi=4 \cdot 4-0 \cdot 4=16 \mathrm{Nm}^2 / \mathrm{C}
\)

Hint

(d) The length of the half-ring is half the circumference of a full circle: \(l=\pi R\).
The total charge \(Q\) is the product of the linear charge density \(\lambda\) and the length \(l\) :
\(Q=\lambda l\)
\(Q=\lambda \pi R\)
\(Q=\left(4 \times 10^{-9} \mathrm{C} / \mathrm{m}\right) \times \pi \times(0.1 \mathrm{~m})\)
\(Q=4 \pi \times 10^{-10} \mathrm{C}\)
The electric potential \(\boldsymbol{V}\) at the center of the half-ring is given by:
\(V=\frac{k Q}{R}\)
\(V=\frac{\left(9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}^2\right)\left(4 \pi \times 10^{-10} \mathrm{C}\right)}{0.1 \mathrm{~m}}\)
\(V=\frac{36 \pi \mathrm{~V} \mathrm{~m}}{0.1 \mathrm{~m}}\)
\(V=360 \pi \times 10^{-1} \mathrm{~V}\)
\(V=36 \pi \mathrm{~V}\)
We are given that \(V=x \pi \mathrm{~V}\).
Comparing this with our calculated value \(V=36 \pi \mathrm{~V}\), we get:
\(x \pi=36 \pi\)
\(x=36\)
The value of \(x\) is 36 .

Hint

(a) 

\(
\begin{aligned}
& E_2 \cos \theta=E_3 \cos \phi \\
& \frac{q_2}{\left(2^2+4^2\right)} \frac{4}{\sqrt{2^2+4^2}}=\frac{q_3}{4^2+3^2} \frac{4}{\sqrt{4^2+3^2}} \\
& \frac{q_2}{q_3}=\frac{20}{25} \frac{\sqrt{20}}{5}=\frac{4}{5} \times \frac{2 \sqrt{5}}{5} \\
& n=5
\end{aligned}
\)

Hint

(b) The electric flux through a surface is given by the formula:
\(
\Phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~A}}=|\overrightarrow{\mathrm{E}}||\overrightarrow{\mathrm{A}}| \cos \theta
\)
where \(\overrightarrow{\mathrm{E}}\) is the electric field, \(\overrightarrow{\mathrm{A}}\) is the area vector (with magnitude equal to the area of the surface and direction perpendicular to the surface, defined by the unit vector \(\hat{n}\) ), and \(\theta\) is the angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{A}}\). However, when using unit vectors to describe the directions of \(\overrightarrow{\mathrm{E}}\) and \(\hat{n}\), the dot product can be used to simplify the calculation as follows:
\(
\Phi=\overrightarrow{\mathrm{E}} \cdot(\overrightarrow{\mathrm{~A}})=(\overrightarrow{\mathrm{E}} \cdot \hat{n}) A
\)
Given that \(\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\) and \(\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)\), and the area \(A=4 \mathrm{~m}^2\), we can substitute them into our formula. Note that since \(\overrightarrow{\mathrm{A}}=A \hat{n}\), the magnitude of the area vector is the area of the surface itself. First, let’s find \(\overrightarrow{\mathrm{E}} \cdot \hat{n}\) :
\(
\overrightarrow{\mathrm{E}} \cdot \hat{n}=\left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)
\)
To compute the dot product, we multiply corresponding components and then add them up:
\(
\begin{aligned}
& \left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)=\frac{1}{6}(2 \cdot 2+6 \cdot 1+8 \cdot 1) \\
& =\frac{1}{6}(4+6+8)=\frac{18}{6}=3
\end{aligned}
\)
Then, the electric flux through the surface is:
\(
\begin{aligned}
& \Phi=(\overrightarrow{\mathrm{E}} \cdot \hat{n}) A=3 \times 4 \mathrm{~m}^2 \\
& \Phi=12 \mathrm{Vm}
\end{aligned}
\)
So, the electric flux for that surface is 12 Vm .

Hint

(b)

\(
\begin{aligned}
\overrightarrow{\mathrm{E}}_{\mathrm{p}} & =\left(\frac{\sigma}{2 \varepsilon_0}+\frac{2 \sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0}\right)(-\hat{\mathrm{i}}) \\
& =-\frac{2 \sigma}{\varepsilon_0} \hat{\mathrm{i}}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& E=\frac{2 k p}{r^3} \\
& E_R=\frac{k p}{(2 r)^3}=\frac{1}{8}\left(\frac{E}{2}\right) \\
& =\frac{E}{16} \\
& \therefore \quad x=16
\end{aligned}
\)

Hint

(d) Electric Field Due to the Infinite Plane Sheet of Charge:
The electric field \(E_s\) due to an infinite plane sheet of charge with surface charge density \(\sigma\) is given by:
\(
E_s=\frac{\sigma}{2 \epsilon_0}
\)
Electric Field Due to the Line Charge:
The electric field \(E_\lambda\) at a perpendicular distance \(r\) from an infinitely long line charge with linear charge density \(\lambda_e\) is:
\(
E_\lambda=\frac{\lambda_e}{2 \pi \epsilon_0 r}
\)
where \(r=4-2=2 \mathrm{~m}\) (the distance from the line charge at \(z=4 \mathrm{~m}\) to the point \((0,0,2)\) ).
Substitute Values and Simplify:
Given \(|\sigma|=2\left|\lambda_e\right|\), we substitute this into the expressions for \(E_s\) and \(E_\lambda\) :
\(
\begin{aligned}
& E_s=\frac{\sigma}{2 \epsilon_0}=\frac{2 \lambda_e}{2 \epsilon_0}=\frac{\lambda_e}{\epsilon_0} \\
& E_\lambda=\frac{\lambda_e}{2 \pi \epsilon_0 \times 2}=\frac{\lambda_e}{4 \pi \epsilon_0}
\end{aligned}
\)
Calculate the Ratio of the Electric Fields:
The ratio of the magnitudes of electric fields \(\frac{E_s}{E_\lambda}\) is:
\(
\frac{E_s}{E_\lambda}=\frac{\frac{\lambda_c}{\epsilon_0}}{\frac{\lambda_c}{4 \pi \varepsilon_0}}=4 \pi
\)
Therefore,
\(
\frac{E_s}{E_\lambda}=\pi \sqrt{16}: 1
\)
Comparing with \(\pi \sqrt{n}: 1\), we find \(n=16\).

Hint

(c)

\(
\begin{aligned}
&\sin \theta=\frac{10}{20}=\frac{1}{2}\\
&\theta=30^{\circ}\\
&\tan \theta=\frac{\mathrm{qE}}{\mathrm{mg}}\\
&\tan 30^{\circ}=\frac{\mathrm{q} \times 2 \times 10^4}{1 \times 10^{-3} \times 10}\\
&\frac{1}{\sqrt{3}}=q \times 10^6\\
&q=\frac{1}{\sqrt{3}} \times 10^{-6} C\\
&x=3
\end{aligned}
\)

Hint

(a)

In air \(\tan \frac{\theta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 r^2 m g}\)
In water \(\tan \frac{\theta}{2}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg}^{\prime}}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{r}^2 \mathrm{mg}_{\text {elf }}}\)
Equate both equations
\(
\varepsilon_0 g=\varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{~g}\left[1-\frac{1}{1.5}\right]
\)
\(
\varepsilon_{\mathrm{r}}=3
\)

Hint

(d)

\(
\begin{aligned}
& \mathrm{V}=\frac{\mathrm{Kp} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right) \\
& =-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2} \\
& \Rightarrow \alpha=27
\end{aligned}
\)

Hint

The angle between the strings is \(37^{\circ}\) in both air and liquid.
Density of the liquid \(\rho_l=0.7 \frac{\mathrm{~g}}{\mathrm{~cm}^3}\).
Density of the sphere \(\rho_s=1.4 \frac{\mathrm{~g}}{\mathrm{~cm}^3}\).
\(
\tan 37^{\circ}=\frac{3}{4}
\)
The electrostatic force is reduced by a factor of the dielectric constant \(K\) in a medium.
Buoyant force \(F_b=V \rho_{l} g\), where \(V\) is the volume of the sphere.
How to solve
Equate the ratio of electrostatic force to the effective weight of the sphere in air and liquid to find the dielectric constant.

Step 1: Forces in air
The electrostatic force is \(F_e\).
The weight of the sphere is \(W=m g=V \rho_s g\).
\(
\tan \frac{\theta}{2}=\frac{F_e}{W}
\)
Step 2: The electrostatic force is \(F_e^{\prime}=\frac{F_e}{K}\).
The effective weight is \(W^{\prime}=W-F_b=V \rho_s g-V \rho_l g=V g\left(\rho_s-\rho_l\right)\).
\(
\tan \frac{\theta}{2}=\frac{F_e^{\prime}}{W^{\prime}}=\frac{F_e}{K V g\left(\rho_s-\rho_l\right)}
\)
Step 3:
\(
\begin{aligned}
& \frac{F_e}{V \rho_s g}=\frac{F_e}{K V g\left(\rho_s-\rho_l\right)} . \\
& \rho_s=K\left(\rho_s-\rho_l\right) .
\end{aligned}
\)
Step 4:
\(
\begin{aligned}
K & =\frac{\rho_s}{\rho_s-\rho_l} . \\
K & =\frac{1.4}{1.4-0.7} . \\
K & =\frac{1.4}{0.7} . \\
K & =2 .
\end{aligned}
\)
The dielectric constant of the liquid is 2.

Hint

(a) The electric field \(E\) due to an infinite plane sheet with surface charge density \(\sigma\) is:
\(E=\frac{\sigma}{2 e_0}\)
The force \(F\) on the electron due to the electric field \(E\) is:
\(F=e E\)
\(F=e \frac{\sigma}{2 \epsilon_0}\)
Using Newton’s second law \(F=m a\), the acceleration \(a\) of the electron is:
\(a=\frac{F}{m}\)
\(\quad a=-\frac{e \sigma}{2 m \epsilon_0}\) (The acceleration is negative because it’s acting in the opposite direction to the initial velocity, causing the electron to decelerate and stop at the sheet)
The kinematic equation for displacement \(S\) is:
\(
S=u t+\frac{1}{2} a t^2
\)
Substitute values:
\(
-1=1 \times 1+\frac{1}{2} \times-\frac{\sigma e}{2 \varepsilon_0 m} \times(1)^2
\)
Solve for \(\sigma\) :
\(
\begin{aligned}
& -1=1-\frac{\sigma e}{4 \varepsilon_0 m} \\
& \frac{\sigma e}{4 \varepsilon_0 m}=2
\end{aligned}
\)
Therefore, \(\sigma=8 \frac{\varepsilon_0 m}{e}\)
Determine \(\alpha\) :
Given \(\sigma=\alpha\left[\frac{m \varepsilon_0}{e}\right]\), we find \(\alpha=8\).

Hint

(b)

\(
\begin{aligned}
& \vec{\tau}=\vec{p} \times \vec{E} \\
& \vec{p}=q \vec{\ell} \\
& \overrightarrow{\mathrm{E}}=0.2 \frac{\mathrm{~V}}{\mathrm{~cm}}=20 \frac{\mathrm{~V}}{\mathrm{~m}} \\
& \overrightarrow{\mathrm{p}}=4 \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& =(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mu \mathrm{C}-\mathrm{m} \\
& \vec{\tau}=(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times(20 \hat{\mathrm{i}}) \times 10^{-6} \mathrm{Nm} \\
& =(8 \hat{\mathrm{k}}+8 \hat{\mathrm{j}}) \times 10^{-5}=8 \sqrt{2} \times 10^{-5} \\
& \alpha=2
\end{aligned}
\)

Hint

(c) Concept:
The total charge on a nucleus is given by:
\(
q=Z \times e
\)
where \(\mathrm{q}=\) charge
\(Z=\) atomic number of the nucleus
\(e=\) charge of electron
Electrical potential is given by:
\(
\mathrm{V}=\frac{1}{4 \pi \epsilon_0} \frac{q}{r}
\)
Calculation:
Given \(\mathrm{z}=50, \mathrm{r}=9 \times 10^{-15} \mathrm{~m}\)
\(
\begin{aligned}
& q=Z \times e \\
& q=50 \times 1.6 \times 10^{-19} \\
& q=8 \times 10^{-18} C
\end{aligned}
\)
Potential at the surface of a nucleus is given as:
\(
\begin{aligned}
& \mathrm{V}=\frac{1}{4 \pi \epsilon_0} \frac{q}{r} \\
& \text { Since, } \frac{1}{4 \pi \epsilon_o}=9 \times 10^9 \\
& \mathrm{~V}=\frac{9 \times 10^9 \times 8 \times 10^{-18}}{9 \times 10^{-15}} \\
& =8 \times 10^6 \mathrm{volt}
\end{aligned}
\)

Hint

(c) 

\(
\begin{aligned}
&\operatorname{Sin}\left(\frac{d \theta}{2}\right) \simeq \frac{d \theta}{2} \text { for very small value of } \theta\\
&2 \mathrm{~T}\left(\frac{d \theta}{2}\right)=\frac{k \lambda R d \theta}{R^2}\\
&\begin{aligned}
\lambda & =\frac{Q}{2 \pi R A} \\
T & =\frac{k Q q}{2 \pi R^2}=3 N
\end{aligned}
\end{aligned}
\)