NEET Practice Q & A

Hint

(a) Density of electric field lines at a point i.e. no. of lines per unit area shows magnitude of electric field at that point.
The electric field lines are denser at A than at B.
Therefore, the electric field is stronger at \(A\) than at \(B\).
\(\circ E_A>E_B\)

Hint

\(
\begin{aligned}
&\text { (b) Electric field on the surface of a conducting sphere is }\\
&\begin{aligned}
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \\
\therefore \text { Charge, } q & =E r^2 \cdot 4 \pi \varepsilon_0=\frac{3 \times 10^6 \times(2.5)^2}{9 \times 10^9} \simeq 2 \times 10^{-3} \mathrm{C}
\end{aligned}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\begin{aligned}
& \mathrm{EA}=\frac{q}{\epsilon_0} \\
& \text { so } \mathrm{E}=\frac{q}{A \epsilon_0}=\frac{\sigma}{\epsilon_0}
\end{aligned}\\
&\text { and normal to the surface }
\end{aligned}
\)

Hint

(d) Electric line of force are perpendicular to the surface of a conductor. Inside the sphere, no lines are present.

Hint

(c) Charge \(q_1=20 \mu \mathrm{C}=20 \times 10^{-6} \mathrm{C}\)
Charge \(q_2=80 \mu \mathrm{C}=80 \times 10^{-6} \mathrm{C}\)
Distance between charges \(d=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Electric field due to a point charge: \(E=k \frac{|q|}{r^2}\), where \(k=8.9875 \times 10^9 \frac{\mathrm{~N} \cdot \mathrm{~m}^2}{\mathrm{C}^2}\)
The electric field is a vector quantity.
The electric field is zero at a point where the magnitudes of the electric fields due to the two charges are equal and opposite in direction.
Let \(x\) be the distance from \(q_1\) where the electric field is zero.
The distance from \(q_2\) is then \(d-x\).
The electric fields are equal in magnitude:
\(k \frac{\left|q_1\right|}{x^2}=k \frac{\left|q_2\right|}{(d-x)^2}\)
\(\frac{\left|q_1\right|}{x^2}=\frac{\left|q_2\right|}{(d-x)^2}\)
Substitute \(q_1=20 \times 10^{-6} \mathrm{C}, q_2=80 \times 10^{-6} \mathrm{C}\), and \(d=0.1 \mathrm{~m}\) into the equation:
\(\frac{20 \times 10^{-6}}{x^2}=\frac{80 \times 10^{-6}}{(0.1-x)^2}\)
\(\frac{1}{x^2}=\frac{4}{(0.1-x)^2}\)
\(
x=\frac{0.1}{3}=0.0333 \mathrm{~m}
\)
The electric field strength will be zero at a distance of 0.033 m from the \(20 \mu \mathrm{C}\) charge.

Hint

(d) For a system of charges to be in equilibrium, the net force on each charge must be zero.
Coulomb’s law: The electrostatic force between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by \(F=k \frac{\left|q_1 q_2\right|}{r^2}\), where \(k\) is Coulomb’s constant.
Let the distance between the two charges \(Q\) be \(2 r\).
The distance between each charge \(Q\) and the charge \(q\) is \(r\).
The force between the two charges \(Q\) is \(F_{Q Q}=k \frac{Q^2}{(2 r)^2}=k \frac{Q^2}{4 r^2}\).
The force between one of the charges \(Q\) and the charge \(q\) is \(F_{Q_q}=k \frac{Q q}{r^2}\).
For the system to be in equilibrium, the net force on one of the charges \(Q\) must be zero.
\(
\begin{aligned}
& F_{Q Q}+F_{Q_q}=0 \\
& k \frac{Q^2}{4 r^2}+k \frac{Q q}{r^2}=0
\end{aligned}
\)
\(
q=-\frac{Q}{4}
\)
The value of \(q\) for which the system is in equilibrium is \(-\frac{Q}{4}\).

Hint

(b) Two point charges: \(q\) and \(2 q\).
Electric field at the location of \(q\) is \(E\).
The force between the charges \(q\) and \(2 q\) separated by a distance \(r\) is given by Coulomb’s law:
\(F=k \frac{|q(2 q)|}{r^2}=k \frac{2 q^2}{r^2}\)
The electric field at the location of charge \(q\) is given as \(E\).
It can also be expressed as the force on charge \(q\) divided by the charge \(q\) :
\(E=\frac{F}{q}=\frac{k \frac{2 q^2}{r^2}}{q}=k \frac{2 q}{r^2}\)
The electric field at the location of charge \(2 q\) can be calculated similarly:
\(E^{\prime}=\frac{F}{2 q}=\frac{k \frac{2 q^2}{r^2}}{2 q}=k \frac{q}{r^2}\)
Divide the expression for \(E^{\prime}\) by the expression for \(E\) :
\(\frac{E^{\prime}}{E}=\frac{k \frac{q}{r^2}}{k \frac{2 q}{r^2}}=\frac{1}{2}\)
\(
E^{\prime}=\frac{1}{2} E
\)
The electric field at the location of the \(2 q\) charge is \(\frac{E}{2}\).

Hint

(a) The electric field at a distance \(\frac{3 R}{2}\) from the center of a charged conducting spherical shell of radius \(R\) is \(E\).
The electric field inside a charged conducting spherical shell is zero.
The electric field outside a charged conducting spherical shell is the same as that of a point charge located at the center of the shell.
The formula for the electric field due to a point charge is \(E=\frac{k Q}{r^2}\), where \(k\) is Coulomb’s constant, \(Q\) is the charge, and \(r\) is the distance from the charge.
The point at a distance \(\frac{R}{2}\) from the center is inside the spherical shell since \(\frac{R}{2}<R\).
The electric field inside a charged conducting spherical shell is zero. Therefore, the electric field at a distance \(\frac{R}{2}\) from the center is 0.
The electric field at a distance \(\frac{R}{2}\) from the center of the sphere is 0.

Hint

(c)

The electric field due to an infinite sheet of charge is \(E=\frac{\sigma}{2 \epsilon_0}\).
The electric field is a vector quantity, so direction matters.
The electric field due to the first sheet is \(E_1=\frac{\sigma}{2 e_0}\), pointing away from the sheet.
The electric field due to the second sheet is \(E_2=\frac{\sigma}{2 \epsilon_0}\), pointing away from the sheet.
Since the point is between the sheets and the charges are alike, the electric fields due to each sheet are in opposite directions.
The net electric field is \(E=E_1-E_2=\frac{\sigma}{2 \epsilon_0}-\frac{\sigma}{2 \epsilon_0}=0\).
The electric field intensity at a point between two parallel sheets with like charges of the same surface charge density is zero.

Hint

\((b)\) Force, \(F=k \frac{q_1 q_2}{r^2} \Rightarrow 12=k \frac{2 \times 6}{r^2} \dots(i)\)
When a charge of -4 C is given to each of these charges, then
\(
\begin{aligned}
q_1 & =-2 \mathrm{C}, q_2=2 \mathrm{C} \\
F^{\prime} & =k \frac{(-2)(2)}{r^2} \dots(ii)
\end{aligned}
\)
On dividing Eq. (ii) by Eq. (i), we get
\(
\begin{aligned}
& \frac{F^{\prime}}{12}=\frac{-4}{12} \\
& F^{\prime}=-4 \mathrm{~N}
\end{aligned}
\)
Here, negative sign indicates that force is attractive.

Hint

(b)

Force, \(F_{12}=\frac{k\left(q^2\right)}{a^2}\)
Force, \(F_{13}=\frac{k\left(q^2\right)}{(\sqrt{2} a)^2}\)
Ratio, \(\frac{F_{12}}{F_{13}}=\frac{2}{1}\)

Hint

(a) Gravitational force,
\(
\begin{aligned}
F_G & =\frac{G m_e m_p}{r^2} \\
F_G & =\frac{6.7 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-27}}{\left(5 \times 10^{-11}\right)^2} \\
& =3.9 \times 10^{-47} \mathrm{~N}
\end{aligned}
\)
Electrostatic force, \(F_e=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}\)
\(
\begin{aligned}
F_e & =\frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(5 \times 10^{-11}\right)^2} \\
& =9.22 \times 10^{-8} \mathrm{~N}
\end{aligned}
\)
So,
\(
\begin{aligned}
\frac{F_e}{F_G} & =\frac{9.22 \times 10^{-8}}{3.9 \times 10^{-47}} \\
& =2.36 \times 10^{39}
\end{aligned}
\)

Hint

(c) The situation is shown in the figure below

Force between \(A\) and \(C\),
\(
F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}
\)
When sphere \(B\) is kept at the mid-point of line joining \(A\) and \(C\), then net force on \(B\) is
\(
\begin{aligned}
F_{\text {net }} & =F_A+F_C=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{(r / 2)^2}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{(r / 2)^2} \\
& =8 \cdot \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}=8 F
\end{aligned}
\)
The net force is towards the \(-q\) charge because the attractive force \(F_C\) is in that direction.
The force experienced by the third charge is \(8 F\) towards the \(-q\) charge

Hint

(b) When the two spheres are brought into contact, the total charge is shared equally because they have the same radius. The total charge \(q\) is:
\(
q=q_1+q_2=10 \mu C+(-20 \mu C)=-10 \mu C
\)
Since both spheres are identical, the charge on each sphere after contact will be:
\(
q_1^{\prime}=q_2^{\prime}=\frac{q}{2}=\frac{-10 \mu C}{2}=-5 \mu C
\)
\(
\frac{F_1}{F_2}=\frac{q_1 q_2}{q_1^{\prime} q_2^{\prime}}=\frac{(10)(-20)}{(-5) \times(-5)}=-\frac{8}{1} \quad\left(\because F \propto q_1 q_2\right)
\)

Hint

(c) Distance between spheres: \(r=1 \mathrm{~m}\)
Attractive force: \(F\)
Repulsive force: \(\frac{F}{3}\)
Coulomb’s Law: \(F=k \frac{\left|q_1 q_2\right|}{r^2}\), where \(k=\frac{1}{4 \pi \varepsilon_0}\)
The attractive force is given by:
\(F=k \frac{\left|q_1\right|\left|-q_2\right|}{r^2}\)
\(F=k \frac{q_1 q_2}{r^2}\)
After connecting the spheres, the charge on each sphere is:
\(q=\frac{q_1+\left(-q_2\right)}{2}=\frac{q_1-q_2}{2}\)
The repulsive force is given by:
\(\frac{F}{3}=k \frac{q^2}{r^2}\)
\(\frac{F}{3}=k \frac{\left(\frac{q_1-q_2}{2}\right)^2}{r^2}\)
\(\frac{F}{3}=k \frac{\left(q_1-q_2\right)^2}{4 r^2}\)
Divide the equation for \(F\) by the equation for \(\frac{F}{3}\) :
\(\frac{F}{\frac{F}{3}}=\frac{k \frac{q_1 q_2}{r^2}}{k \frac{\left(q_1-q_2\right)^2}{4 r^2}}\)
\(3=\frac{4 q_1 q_2}{\left(q_1-q_2\right)^2}\)
Let \(x=\frac{q_1}{q_2}\)
\(
x=\frac{10 \pm 8}{6}
\)
The ratio of the magnitudes of the initial charges is \(3: 1\) or \(1: 3\).

Hint

(b)

Electric field due to charge \(q\) at the center
\(
E_A=\frac{k q}{\left(\frac{a}{\sqrt{2}}\right)^2}=\frac{2 k q}{a^2}=E
\)
Electric field at the center due to charge \(2 q\)
\(
E_B=2\frac{k q}{\left(\frac{a}{\sqrt{2}}\right)^2}=2\frac{2 k q}{a^2}=2E
\)
Similarly \(E_C=3 E, E_D=4 E\)
Resultant of \(E_C\) and \(E_A\) will be in the direction of \(E_C\) because its value is greater
The resultant of \(E_B\) and \(E_D\) is in the direction of \(E_D\) because its value is greater
Now, from the figure. We can say that the results of the electric field in the direction of \(E_C\) and \(E_D\) will be along \(C B\)
So, the resultant is along \(C B\).

Hint

(c) We have a ball with a charge of \(-50 e\) placed at the center of a hollow spherical shell. The shell itself has a net charge of \(-50 e\).
According to Gauss’s Law, the electric field inside a conductor in electrostatic equilibrium is zero. This means that any excess charge on the shell will redistribute itself in such a way that the electric field inside the shell (and thus at the location of the charge) is zero.
The charge on the ball \(-50 e\) will induce a charge of \(+50 e\) on the inner surface of the shell. This is because the inner surface must have a charge that cancels the electric field due to the charge at the center to maintain zero electric field inside the conductor.
The total charge on the shell is \(-50 e\). Since we have induced \(+50 e\) on the inner surface, we can find the charge on the outer surface by using the equation:
Charge on outer surface \(=\) Total charge on shell – Charge on inner surface
Substituting the values:
Charge on outer surface \(=-50 e-(+50 e)=-50 e\)
\(
-50 e=-100 e
\)
The charge on the outer surface of the shell is \(-\mathbf{1 0 0 e}\).

Hint

(c) Electric field in vacuum, \(E_0=\frac{\sigma}{\varepsilon_0}\)
In medium, \(E_2=\frac{\sigma}{\varepsilon_0 K}\)
If \(K>1[latex], then [latex]E_2<E_0\).
i.e., if the plates are dipped in kerosene oil tank, the electric field between the plates will decrease.

Note: The dielectric constant of kerosene oil is greater than 1 (\(\)K>1\(\)).

Hint

(a) The force due to the electric field is \(F_e=q E\).
The friction force is \(F_f=\mu N\).
The normal force on a horizontal surface is \(N=m g\).
To find the coefficient of friction when the kinetic energy of the block remains constant, we need to consider the forces acting on the block. The frictional force must be balanced by the electric force acting on the charged block. The kinetic energy (KE) of the block is given by \(K E=\frac{1}{2} m v_0^2\). Since the kinetic energy remains constant, the work done by the frictional force must equal the work done by the electric force.
Setting these two forces equal gives us: \(\mu m g=q E\). Solving for \(\mu[latex] gives us:(where, [latex]\mu[latex] is coefficient of friction)
[latex]
\therefore \quad \mu=\frac{q E}{m g}
\)

Hint

(a)

Solid sphere radius: \(a\)
Solid sphere charge: \(2 Q\)
Spherical shell inner radius: \(b\)
Spherical shell outer radius: \(c\)
Spherical shell charge: \(-Q\)
For a conductor in electrostatic equilibrium, the electric field inside the conductor is zero.
Surface charge density \((\sigma)=\frac{\text { Charge }}{\text { Surface area }}\)
By induction, the charge on the inner surface of the shell is equal and opposite to the charge on the solid sphere
The charge on the inner surface of the shell is \(-2 Q\).
The net charge on the shell is \(-Q\).
The charge on the outer surface of the shell is the net charge minus the charge on the inner surface: \(-Q-(-2 Q)=Q\).
The area of the inner surface is \(4 \pi b^2\).
The surface charge density on the inner surface is \(\sigma_{\text {inner }}=\frac{-2 Q}{4 \pi b^2}=-\frac{Q}{2 \pi b^2}\).
The area of the outer surface is \(4 \pi c^2\).
The surface charge density on the outer surface is \(\sigma_{\text {outer }}=\frac{Q}{4 \pi c^2}\).
The surface charge density on the inner surface of the shell is \(-\frac{Q}{2 \pi b^2}\) and the surface charge density on the outer surface of the shell is \(\frac{Q}{4 \pi c^2}\).

Hint

(d) Gauss’s law of electrostatic states that, the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity, i.e. \(\phi=\frac{Q_{enc}}{\varepsilon_0}\). Thus, electric flux through a surface does not depend on the shape, size or area of a surface but it depends on the total charge enclosed by the surface.
So, here in this question, all the figures have same electric flux as all of them has single positive charge.

Hint

(b) When charged particle enters in a uniform electric field, then force on charged particle, \(F=q E\)
Also, \(F=m a\)
\(
\therefore \quad m a=q E
\)
or acceleration, \(a=\frac{q E}{m}=\frac{3 \times 10^{-3} \times 80}{20 \times 10^{-3}}=12 \mathrm{~ms}^{-2}\)
So, from equations of motion
\(
v=u+a t=20+12 \times 3=56 \mathrm{~ms}^{-1}
\)

Hint

(b) On the outer surfaces of the shell surface charge densities are equal

\(
\frac{Q_3+Q_2+Q_1}{4 \pi(3 R)^2}=\frac{Q_2+Q_1}{4 \pi(2 R)^2}=\frac{Q_1}{4 \pi R^2}
\)
\(
\frac{Q_3+Q_2+Q_1}{9}=\frac{Q_2+Q_1}{4} \text { and } \frac{Q_2+Q_1}{4}=\frac{Q_1}{1}
\)
\(
\begin{aligned}
& \Rightarrow \quad Q_2=3 Q_1 \\
& \therefore \quad \frac{Q_3+Q_2+Q_1}{9}=\frac{Q_1}{1} \Rightarrow Q_3=5 Q_1 \\
& \text { Hence, } Q_1: Q_2: Q_3=1: 3: 5
\end{aligned}
\)

Hint

(c) The situation is shown below.

The direction of the dipole moment is shown in the figure.

Net electric dipole moment,
\(
\begin{aligned}
p_{\text {net }} & =\sqrt{p^2+p^2+2 p p \cos 60^{\circ}}=\sqrt{3} p \\
= & \sqrt{3} q l \quad[\because p=q l]
\end{aligned}
\)

Hint

(d) Electric field at centre due to circular portion is zero because electric field due to each charged element at centre will be cancel out by the electric field due the charged element of its just opposite side.

Hint

(a) According to Gauss’s theorem, the electric flux through any closed surface is equal to \(1 / \varepsilon_0\) times the net charge \(q\) enclosed by the surface.
\(
\text { Flux of electric field, } \phi=\frac{1}{\varepsilon_0} \times Q_{\text {enc }}=\frac{1}{\varepsilon_0}(2 q)
\)

Hint

(d)

Flux through surface \(A, \phi_A=E \pi R^2 \operatorname{Cos} 180^{\circ}=\mathrm{E} \pi \mathrm{R}^2 \operatorname{Cos} 0^{\circ}=\mathrm{E} \pi R^2\)
Flux through curved surface \(C\),
\(
\phi_C=\int \mathbf{E} \cdot d \mathbf{S}=\int E d S \cos 90^{\circ}=0
\)
\(\therefore\) Total flux through cylinder \(=\phi_A+\phi_B+\phi_C=0\)

Hint

(d) Electric flux,
\(
\begin{aligned}
\phi_E & =\int \mathbf{E} \cdot d \mathbf{S} \\
& =\int E d S \cos \theta=\int E d S \cos 90^{\circ}=0
\end{aligned}
\)
Note: Area vector is perpendicular to Electric field vector.

Hint

(c) Let the initial charges on the spheres be \(q_1\) and \(q_2\).
Since they attract, one charge must be positive and the other negative.
Let \(q_1=\)+ve and \(q_2=\)-ve.
The initial attractive force is \(F_1=k \frac{q_1 q_2}{d^2}\).
When the spheres are brought into contact, the total charge is \(q_1-q_2\).
This total charge is shared equally between the two spheres, so each sphere has a charge of \(\frac{q_1-q_2}{2}\).
\(F_2=\frac{k\left(\frac{Q_1-Q_2}{2}\right)^2}{d^2}\)
\(
\begin{aligned}
&\text { According to question, } F_1=F_2\\
&\begin{aligned}
Q_1 Q_2 & =\frac{\left(Q_1-Q_2\right)^2}{4} \\
\Rightarrow \quad 4 Q_1 Q_2 & =Q_1^2+Q_2^2-2 Q_1 Q_2 \\
0 & =Q_1^2+Q_2^2-6 Q_1 Q_2 \\
\Rightarrow \quad \frac{Q_1}{Q_2} & =3 \pm \sqrt{8}
\end{aligned}
\end{aligned}
\)

Hint

(c) The acceleration of the electron due to given coulombic force, \(F\) is \(a_e=\frac{F}{m_e} \dots(i)\)
where, \(m_e\) is the mass of the electron.
The acceleration of the proton due to same force \(F\) is
\(
a_p=\frac{F}{m_p} \dots(ii)
\)
where, \(m_p\) is the mass of the proton.
On dividing Eq. (ii) by Eq. (i), we get \(\frac{a_p}{a_e}=\frac{m_e}{m_p}\)
\(
\begin{aligned}
a_p & =\frac{a_e m_e}{m_p}=\frac{\left(2.5 \times 10^{22} \mathrm{~ms}^{-2}\right)\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}{\left(1.67 \times 10^{-27} \mathrm{~kg}\right)} \\
& =13.6 \times 10^{18} \mathrm{~ms}^{-2} \approx 1.5 \times 10^{19} \mathrm{~ms}^{-2}
\end{aligned}
\)

Hint

(a) By using, \(Q E=m g\) \(\Rightarrow E=\frac{m g}{Q}=\frac{10^{-6} \times 10}{10^{-6}}=10 \mathrm{~V} / \mathrm{m}\), upward because charge is positive.

Note: For the drop to be balanced, the electric force must equal the weight:
\(F=W\)
\(q E=m g\)

Hint

(b) In Figs. (1), (3), and (4),the  net electric field is zero because the electric field at a point due to a positive charge acts away from the charge, and due toa  negative charge, it acts towards the charge. So for Fig. (2),

Here, the net electric field in Fig. (2) is
\(
=\sqrt{(2 E)^2+(2 E)^2+(2 E)(2 E) \cdot 2 \cos 120^{\circ}}=2 E
\)
which is not zero.

Hint

(a) According to Gauss’s law, total flux coming out of a closed surface enclosing charge \(q\) is given by \(\phi=\int \mathbf{E} \cdot d \mathbf{S}=\frac{q}{\varepsilon_0}\).
From this expression, it is clear that total flux linked with a closed surface only depends on the enclosed charge and independent of the shape and size of the surface.
\(
\phi=\int \mathbf{E} \cdot d \mathbf{S}=\frac{q}{\varepsilon_0}=20 \mathrm{Vm} \quad \quad \text { [given] }
\)
This \(\frac{q}{\varepsilon_0}\) is constant as long as the enclosed charge is constant.
The flux over a concentric sphere of radius \(20 \mathrm{~cm}=20 \mathrm{~V}\mathrm{m}\).

Hint

(b) Given, \(A(1,0,4)\) and \(B(2,-1,5)\)
\(
\begin{array}{ll}
\therefore \quad & \mathbf{A B}=[(2-1) \hat{\mathbf{i}}+(-1-0) \hat{\mathbf{j}}+(5-4) \hat{\mathbf{k}}] \\
& \mathbf{A B}=[\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}]
\end{array}
\)
Torque, \(\tau=\mathbf{p} \times \mathbf{E}=q \mathbf{A B} \times \mathbf{E}\)
\(
\begin{aligned}
\tau & =4 \times 10^{-6}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times 20 \hat{\mathbf{i}} \\
\tau & =8 \times 10^{-5}(\hat{\mathbf{k}}+\hat{\mathbf{j}})
\end{aligned}
\)
Magnitude of torque,
\(
\begin{aligned}
\tau & =8 \times 10^{-5} \sqrt{1^2+1^2} \\
\tau & =1.13 \times 10^{-4} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)

Hint

(b) Given, \(E=7.18 \times 10^8 \mathrm{~N} / \mathrm{C}\)
and \(\quad r=2 \mathrm{~cm}=2 \times 10^{-2}\)
Electric field is given by
\(
\begin{aligned}
E & =\frac{\lambda}{2 \pi \varepsilon_0 r} \\
\lambda & =2 \pi \varepsilon_0 r E \\
\lambda & =\frac{4 \pi \varepsilon_0 r E}{2}=\frac{2 \times 10^{-2} \times 7.18 \times 10^8}{2 \times 9 \times 10^9} \\
& =7.98 \times 10^{-4} \mathrm{C} / \mathrm{m}
\end{aligned}
\)

Hint

\(
\text { (a) Here, } p=\left(500 \times 10^{-6}\right) \times\left(10 \times 10^{-2}\right)=5 \times 10^{-5} \mathrm{C}-\mathrm{m}
\)

\(
\begin{aligned}
&r=25 \mathrm{~cm}=0.25 \mathrm{~m}, a=5 \mathrm{~cm}=0.05 \mathrm{~m}\\
&\text { Electric field intensity, } E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 p r}{\left(r^2-l^2\right)^2}\\
&\begin{aligned}
E & =\frac{9 \times 10^9 \times 2 \times 5 \times 10^{-5} \times 0.25}{\left\{(0.25)^2-(0.05)^2\right\}^2} \\
& =6.25 \times 10^7 \mathrm{~N} / \mathrm{C}
\end{aligned}
\end{aligned}
\)

Hint

(a) Suppose neutral point \(N\) lies at a distance \(x\) from dipole of moment \(p\) or at a distance \((25-x)\) from dipole of moment 64 p.

At point \(N[latex] electric field due to dipole (1) = Electric field due to dipole (2)
[latex]
\begin{aligned}
\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 p}{x^3} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2(64 p)}{(25-x)^3} \\
\frac{1}{x^3} & =\frac{64}{(25-x)^3} \\
x & =5 \mathrm{~cm}
\end{aligned}
\)

Hint

\(\text { d) Initially force, } F=k \frac{Q^2}{r^2} \) from Fig(a)

Finally, when a third spherical conductor cames in contact alternately with \(B\) and \(C\) then removed, the charges on \(B\) and \(C\) become \(Q / 2\) and \(3 Q / 4\), respectively (Fig(b)).
\(
\text { Now, force, } F^{\prime}=k \frac{(Q / 2)(3 Q / 4)}{r^2}=\frac{3}{8} F
\)

Hint

(c) The torque on a dipole moment is \(\tau=\mathbf{p} \times \mathbf{E}\). The torque has maximum value when \(\mathbf{p}\) and \(\mathbf{E}\) are perpendicular to each other, so that \(\sin \theta\) is maximum, i.e. \(\sin \theta=1\).
\(
\begin{aligned}
\tau & =\left(3 \times 10^4\right)\left(6 \times 10^{-30}\right) \\
& =18 \times 10^{-26} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)

Hint

(c) Two positive ions, each carrying a charge \(q\) are kept at a distance \(d\), then it is found that force of repulsion between them is \(F=\frac{k q q}{d^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q q}{d^2}\)
where,
\(
q=n e
\)
\(
\therefore \quad F=\frac{1}{4 \pi \varepsilon_0} \frac{n^2 e^2}{d^2}
\)
Number of electrons, \(n=\sqrt{\frac{4 \pi \varepsilon_0 F d^2}{e^2}}\)

Hint

(c) When both discs \(A\) and \(B\) are touched, charge flows from higher value (higher potential) to lower value (lower potential) till it equalises on the two discs.
Given, \(q_1=10^{-6} \mathrm{C}, q_2=10^{-5} \mathrm{C}\)
\(
\therefore \quad q=\frac{q_1+q_2}{2}=\frac{10^{-6}+10^{-5}}{2}=5.5 \mu \mathrm{C}
\)

Hint

(c) Suppose the field vanishes at a distance \(x\), we have

\(
\begin{gathered}
\frac{k q}{x^2}=\frac{k q / 2}{(x-a)^2} \\
\text { or } \quad 2(x-a)^2=x^2 \text { or } \sqrt{2}(x-a)=x \\
\text { or } \quad(\sqrt{2}-1) x=\sqrt{2} a \text { or } x=\frac{\sqrt{2} a}{\sqrt{2}-1}
\end{gathered}
\)

Hint

(b) Electric field due to \(Q_2\) and \(Q_3\) cancel each other.
So, the vector numbered 2 coincides in direction with the electric field due to \(Q_1\) at mid-point \(M\) of the hypotenuse.

Hint

(c) Three vectors of equal magnitude are inclined at \(120^{\circ}\) with the adjacent vector. So, net electric intensity will be zero.
Note: The electric fields due to the three charges are directed along the lines joining the charges to the centroid, pointing outwards.
Due to the symmetry of the equilateral triangle, the angle between any two electric field vectors is \(\frac{360^{\circ}}{3}=120^{\circ}\).
The vector sum of three equal vectors at \(120^{\circ}\) to each other is zero.
Therefore, the net electric field at the centroid is zero.

Hint

(c) The given situation is shown below.

Here, \(F_A=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{a^2}\) and \(F_C=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{a^2}\)
Net force on \(B, F_{\text {net }}=F_{A C}+F_D\)
\(
=\sqrt{F_A^2+F_C^2}+F_D
\)
\(
\begin{aligned}
& =\sqrt{\left\{\left(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}\right)^2+\left(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}\right)^2\right\}}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{(a \sqrt{2})^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{a^2}\left(\frac{1+2 \sqrt{2}}{2}\right)
\end{aligned}
\)

Hint

(b) If the ring is complete, then the net field at the centre is zero. As small portion is cut, then field opposite to this portion is not cancelled out.
Charge of small portion \(=\lambda \cdot l\)
\(\therefore\) Electric field due to this portion \(=\frac{\lambda l}{4 \pi \varepsilon_0 a^2}\)
\(
\begin{aligned}
&\text { Let } E_R \text { is electric field of remaining portion. }\\
&\begin{array}{ll}
\therefore & E_R+\frac{\lambda l}{4 \pi \varepsilon_0 a^2}=0 \\
\Rightarrow & E_R=\frac{-\lambda l}{4 \pi \varepsilon_0 a^2}
\end{array}
\end{aligned}
\)

Hint

\(
\text { (b) } E \propto \frac{q}{r^2} \Rightarrow \tan \theta=\frac{E_0}{E_0 / 2}=2
\)

Hint

\(
\begin{aligned}
&\text { (b) In the following figure, in equilibrium, }\\
&F_e=T \sin 30^{\circ} \text { and } r=1 \mathrm{~m}
\end{aligned}
\)

\(
\begin{aligned}
&\Rightarrow 9 \times 10^9 \cdot \frac{Q^2}{r^2}=T \times \frac{1}{2} \Rightarrow 9 \times 10^9 \cdot \frac{\left(10 \times 10^{-6}\right)^2}{1^2}=T \times \frac{1}{2}\\
&\therefore \text { Tension in the threads, } T=1.8 \mathrm{~N}
\end{aligned}
\)

Hint

(d) The schematic diagram of distribution of charges on X-axis is shown in the figure below.

Total force acting on 1 C charge is given by
\(
\begin{aligned}
F=\frac{1}{4 \pi \varepsilon_0}\left[\frac{1 \times 1 \times 10^{-6}}{(1)^2}\right. & +\frac{1 \times 1 \times 10^{-6}}{(2)^2} \\
& \left.+\frac{1 \times 1 \times 10^{-6}}{(4)^2}+\frac{1 \times 1 \times 10^{-6}}{(8)^2}+\ldots \infty\right]
\end{aligned}
\)
\(
=\frac{10^{-6}}{4 \pi \varepsilon_0}\left(\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots \infty\right)=9 \times 10^9 \times 10^{-6}\left(\frac{1}{1-(1 / 4)}\right)
\)
\(
=9 \times 10^9 \times 10^{-6} \times \frac{4}{3}=9 \times \frac{4}{3} \times 10^3=12000 \mathrm{~N}
\)

Hint

\(
\begin{aligned}
&\text { (c) Electric force, } q E=m a \Rightarrow a=\frac{q E}{m}\\
&\begin{aligned}
& \therefore \quad a=\frac{1.6 \times 10^{-19} \times 1 \times 10^3}{9 \times 10^{-31}}=\frac{1.6}{9} \times 10^{15} \\
& \because \quad u=5 \times 10^6 \mathrm{~ms}^{-1} \text { and } v=0 \\
& \therefore \text { From } v^2=u^2-2 a s \Rightarrow s=\frac{u^2}{2 a} \\
& \therefore \text { Distance, } s=\frac{\left(5 \times 10^6\right)^2 \times 9}{2 \times 1.6 \times 10^{15}}=7 \mathrm{~cm} \text { (approx.) }
\end{aligned}
\end{aligned}
\)

Hint

(d)

The magnitude of the electric field at an axial point \(P\) at a distance \(z\) from the origin is given by
\(
\left|\mathbf{E}_{(z)}\right|=\frac{4 q a z}{4 \pi \varepsilon_0\left(z^2-a^2\right)^2}=\frac{2 p z}{4 \pi \varepsilon_0\left(z^2-a^2\right)^2}
\)
where, \(p=2 q a\) is the electric dipole moment
For \(z \gg a,\left|\mathbf{E}_{(z)}\right|=\frac{2 p}{4 \pi \varepsilon_0 z^3}\)
The magnitude of electric field at an equatorial point \(Q\) at a distance \(y\) from the origin is given by
\(
\left|\mathbf{E}_{(y)}\right|=\frac{1}{4 \pi \varepsilon_0} \frac{2 q a}{\left(y^2+a^2\right)^{3 / 2}}=\frac{p}{4 \pi \varepsilon_0\left(y^2+a^2\right)^{3 / 2}}
\)
For \(y \gg a,\left|\mathbf{E}_{(y)}\right|=\frac{p}{4 \pi \varepsilon_0 y^3}\)
For
\(
z=y \gg a,
\)
\(
\therefore \quad \frac{\left|\mathbf{E}_{(z)}\right|}{\left|\mathbf{E}_{(y)}\right|}=2
\)

Hint

Hint

(c) According to Gauss’s theorem, electric flux through the sphere \(=\frac{q}{\varepsilon_0}\).
\(\therefore\) Electric flux through the hemisphere \(=\frac{1}{2} \frac{q}{\varepsilon_0}\)
\(
\begin{aligned}
& =\frac{10 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}}=0.56 \times 10^6 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1} \\
& \approx 0.6 \times 10^6 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}=6 \times 10^5 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}
\end{aligned}
\)

Hint

(b) Charge enclosed by cylindrical surface (length 100 cm ) is \(Q_{\text {encl }}=100 Q\). By applying Gauss’s law,
\(
\phi=\frac{1}{\varepsilon_0}\left(Q_{\mathrm{encl}}\right)=\frac{1}{\varepsilon_0}(100 Q)
\)
\(\because\) The lines are parallel to the surface.

Hint

(a) Electric field due to a hollow spherical conductor is given by following equations \(E=0\), for \(r<R\) and
\(
E=\frac{Q}{4 \pi \varepsilon_0 r^2}, \text { for } r \geq R
\)
i.e. inside the conductor field will be zero and outside the conductor will vary according to \(E \propto \frac{1}{r^2}\).

Hint

(c) The field increases linearly from centre inside the sphere (from \(r=0\) to \(r=R\) ) become maximum at the surface of the sphere and decrease rapidly with distance \(\left(\propto \frac{1}{r^2}\right)\) outside the sphere.

Hint

(a) When dipole is given a small angular displacement \(\theta\) about it’s equilibrium position, then the restoring torque will be
\(\tau=-p E \sin \theta=-p E \theta[latex] (as [latex]\sin \theta=\theta\) )
or \(\quad I \frac{d^2 \theta}{d t^2}=-p E \theta\)
\(
\left(\text { as } \tau=I \alpha=I \frac{d^2 \theta}{d t^2}\right)
\)
As, \(\quad \frac{d^2 \theta}{d t^2}=-\omega^2 \theta \Rightarrow \omega^2=\frac{p E}{I} \Rightarrow \omega=\sqrt{\frac{p E}{I}}\)

Hint

(b) Suppose electric field is zero at a point \(P\) lies at a distance \(d\) from the charge \(+Q\).
\(
\begin{aligned}
& \text { At } P, \quad \frac{K Q}{d^2}=\frac{K(2 Q)}{(a+d)^2} \\
& \Rightarrow \quad \frac{1}{d^2}=\frac{2}{(a+d)^2} \Rightarrow d=\frac{a}{(\sqrt{2}-1)}
\end{aligned}
\)

Since, \(d>a\), i.e. point \(P\) must lies on negative \(X\)-axis as shown at a distance \(x\) from origin, hence, \(x=d-a=\frac{a}{(\sqrt{2}-1)}-a=\sqrt{2} a\).
Actually, \(P\) lies on negative \(X\)-axis, so \(x=-\sqrt{2} a\)

Hint

\(
\begin{aligned}
&\text { (a) } \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}=m r \omega^2=\frac{4 \pi^2 m r}{T^2}\\
&\left(\because \omega=\frac{2 \pi}{T}\right)
\end{aligned}
\)
\(
T^2=\frac{\left(4 \pi \varepsilon_0\right) r^2\left(4 \pi^2 m r\right)}{q_1 q_2}
\)
\(
\therefore \text { Time period, } T=\left[\frac{4 \pi^2 m r^3}{k q_1 q_2}\right]^{1 / 2}
\)

Hint

(b) The system is in equilibrium means the force experienced by each charge is zero. It is clear that charge placed at centre would be in equilibrium for any value of \(q\), so we are considering the equilibrium of charge placed at any corner.

\(
F_{C D}+F_{C A} \cos 45^{\circ}+F_{C O} \cos 45^{\circ}=0
\)
\(
\begin{aligned}
\Rightarrow \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(-Q)(-Q)}{a^2}+\frac{1}{4 \pi \varepsilon_0} \frac{(-Q)(-Q)}{(\sqrt{2} a)^2} & \times \frac{1}{\sqrt{2}} \\
& +\frac{1}{4 \pi \varepsilon_0} \frac{(-Q) q}{(\sqrt{2} a / 2)^2} \times \frac{1}{\sqrt{2}}=0
\end{aligned}
\)
\(
\Rightarrow \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{a^2}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{2 a^2} \cdot \frac{1}{\sqrt{2}}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 Q q}{a^2} \cdot \frac{1}{\sqrt{2}}=0
\)
\(
\Rightarrow \quad Q+\frac{Q}{2 \sqrt{2}}-\sqrt{2} q=0 \Rightarrow \quad 2 \sqrt{2} Q+Q-4 q=0
\)
\(
\begin{array}{ll}
\Rightarrow & 4 q=(2 \sqrt{2}+1) Q \\
\text { Therefore, } & q=\frac{Q}{4}(2 \sqrt{2}+1)
\end{array}
\)

Hint

(a) Let the electric field is zero at point \(O\) in the figure.

\(
\begin{aligned}
&\begin{array}{ll}
\therefore & E_1=E_2 \\
\therefore & E_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{a^2} \\
& E_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{b^2}
\end{array}\\
&\text { Also, } x=a+b \text { or } 11=a+b\\
&\begin{aligned}
& \therefore \quad b=11-a \\
& \text { Now, } \quad \frac{1}{4 \pi \varepsilon_0} \frac{q_1}{a^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{(11-a)^2}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
\frac{q_1}{q_2} & =\frac{a^2}{(11-a)^2} \\
\sqrt{\frac{q_1}{q_2}} & =\frac{a}{11-a} \text { or } \sqrt{\frac{25}{36}}=\frac{a}{11-a} \\
\frac{5}{6} & =\frac{a}{11-a} \text { or } 6 a=55-5 a \\
a & =5 \mathrm{~cm}
\end{aligned}
\)
So, intensity will be zero at a distance of 5 cm from \(25 \mu \mathrm{C}\).

Hint

(d) 1 electron has a charge of \(1.6 \times 10^{-19} \mathrm{C}\).
\(10^{10}\) electrons would have a charge of
\(
\begin{aligned}
q & =n e=1.6 \times 10^{-19} \times 10^{10} \\
& =1.6 \times 10^{-9} \mathrm{C}
\end{aligned}
\)
Thus, in 1 s charge accumulated \(=1.6 \times 10^{-9} \mathrm{C}\) So, time taken to accumulate 1 C
\(
\begin{aligned}
& =\frac{1}{1.6 \times 10^{-9}}=0.625 \times 10^9 \\
& =6.25 \times 10^8 \mathrm{~s}=173611 \mathrm{~h} \\
& =7233 \text { days } \approx 20 \mathrm{yr}
\end{aligned}
\)

Hint

(c) The resultant of \(E_A[latex] and [latex]E_B\) is

\(
E_{A B}=\sqrt{E_A^2+E_B^2+2 E_A E_B \cos \left(120^{\circ}\right)}=E_A
\)
Since \(E_A=E_B=E_C[latex], the direction of [latex]E_{A B}\) is opposite to \(E_C\).
The resultant electric field is \(E_{\text {total }}=E_{A B}+E_C=E_A-E_A=0\).

Hint

When sphere \(C\) is touched to \(A\), then equal charge \(Q / 2\) distributes on \(A\) and \(C\).
\(
\therefore \quad F_A=\frac{1}{4 \pi \varepsilon_0} \frac{(Q / 2)^2}{(r / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{r^2}
\)
\(
\begin{aligned}
&F_B=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Q)(Q / 2)}{(r / 2)^2}=2 \cdot \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{r^2}\\
&\therefore \text { Net force on } C, F_{\text {net }}=F_B-F_A=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{r^2}=F
\end{aligned}
\)

Hint

(b) The electric field at \(C\) due to charge \(+10^{-7} \mathrm{C}\) at \(A\) is
\(
\mathbf{E}_1=\frac{1}{4 \pi \varepsilon_0} \frac{10^{-7}}{(0.2)^2} \text { along } A C
\)
The electric field at \(C\) due to charge \(-10^{-7} \mathrm{C}\) at \(B\) is
As,
\(
\begin{aligned}
& \mathbf{E}_2=\frac{1}{4 \pi \varepsilon_0} \frac{10^{-7}}{(0.2)^2} \text { along } C B \\
& \left|\mathbf{E}_1\right|=\left|\mathbf{E}_2\right|
\end{aligned}
\)
By symmetry, the vertical components will cancel out and horizontal components will add.

\(
\begin{aligned}
&\therefore \text { The resultant electric field at } C \text { is }\\
&\begin{aligned}
E & =2 E_1 \cos 60^{\circ}=2 \times \frac{1}{4 \pi \varepsilon_0} \frac{10^{-7}}{(0.2)^2} \times \frac{1}{2} \\
& =\frac{9 \times 10^9 \times 10^{-7}}{(0.2)^2}=2.2 \times 10^4 \mathrm{~N} / \mathrm{C}
\end{aligned}
\end{aligned}
\)

Hint

(c) Number of electrons,
\(
n=\frac{6 \times 10^{23}}{63.5} \times 10 \times \frac{1}{10^6}=\frac{6 \times 10^{18}}{63.5}
\)
\(
\begin{aligned}
q & =n e \\
q & =\frac{6 \times 10^{18} \times 1.6 \times 10^{-19}}{63.5} \\
q & =1.5 \times 10^{-2} \mathrm{C} \\
F & =\frac{9 \times 10^9 \times 1.5 \times 10^{-2} \times 1.5 \times 10^{-2}}{\left(\frac{10}{100}\right)^2} \\
& =2.0 \times 10^8 \mathrm{~N}
\end{aligned}
\)

Hint

(a) The force due to the spring is given by \(F_s=-k x\), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position.
The force due to the electric field is given by \(F_e=Q E\).
At equilibrium, the net force is zero: \(F_s+F_e=0\).
\(
\begin{aligned}
& -k x_{e q}+Q E=0 . \\
& x_{e q}=\frac{Q E}{k} .
\end{aligned}
\)
The net force is \(F=F_s+F_e=m \frac{d^2 x}{d t^2}\).
\(
-k\left(x-x_{e q}\right)=m \frac{d^2 x}{d t^2} .
\)
Let \(y=x-x_{e q}\).
\(
-k y=m \frac{d^2 y}{d t^2} .
\)
\(
\frac{d^2 y}{d t^2}+\frac{k}{m} y=0 .
\)
The angular frequency is \(\omega=\sqrt{\frac{k}{m}}\).
The frequency is \(\nu=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\). Since \(\nu_0=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\), the frequency remains \(\nu_0\).
The frequency will be same, \(\nu=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\) but due to the constant force i.e. \(q E\), the equilibrium position gets shifted by \(\frac{q E}{k}\) in forward direction. 

Hint

(d) The electric field at the center of a uniformly charged ring is zero.
\(
E_{\text {total }}=0
\)
The total charge is the sum of the charges on parts \(A K B\) and \(A C D B\).
\(
\begin{aligned}
& E_{\text {total }}=E_{A K B}+E_{A C D B} \\
& 0=E+E_{A C D B} \\
& E_{A C D B}=-E
\end{aligned}
\)
The electric field at the center due to the charge on the part \(A C D B\) of the ring is \(-E\).
Hence, electric field at \(O\) due to the part \(A C D B\) is equal in magnitude and opposite in direction that due to the part \(A K B\), i.e. \(E\) along \(O K\).

Hint

\(
\begin{aligned}
& \text { (a) Total flux, } \phi_{\text {total }}=\phi_A+\phi_B+\phi_C=\frac{q}{\varepsilon_0} \\
& \because \quad \phi_B=\phi \text { and } \phi_A=\phi_C=\phi^{\prime} \text { [assumed] } \\
& \therefore \quad 2 \phi^{\prime}+\phi=\frac{q}{\varepsilon_0}
\end{aligned}
\)
The flux through the plane surface \(A, \phi^{\prime}=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)\)

Hint

(a) According to Gauss’s theorem,
\(
\begin{aligned}
\oint \mathbf{E} \cdot d \mathbf{S} & =\frac{1}{\varepsilon_0} Q_{\text {enclosed }} \\
E \cdot 4 \pi x^2 & =\frac{Q}{\varepsilon_0}
\end{aligned}
\)
Electric field at the point \(P, E=\frac{Q}{4 \pi \varepsilon_0 x^2}\)

Hint

(a) Charge density of long wire, \(\lambda=\frac{1}{3} \mathrm{C}-\mathrm{m}^{-1}\) From Gauss’s theorem, \(\oint \mathbf{E} \cdot d \mathbf{S}=\frac{q}{\varepsilon_0}\)

\(
\Rightarrow \quad E \oint d S=\frac{q}{\varepsilon_0} \text { or } E 2 \pi r l=\frac{q}{\varepsilon_0}
\)
\(
\begin{aligned}
&\Rightarrow \quad E=\frac{q}{2 \pi \varepsilon_0 r l}=\frac{q / l}{2 \pi \varepsilon_0 r} \Rightarrow \frac{\lambda \times 2}{2 \pi \varepsilon_0 r \times 2}=\frac{\lambda \times 2}{4 \pi \varepsilon_0 r}\\
&\text { The magnitude of the electric intensity, }\\
&\begin{aligned}
E & =9 \times 10^9 \times \frac{1}{3} \times 2 \times \frac{1}{18 \times 10^{-2}} \\
& =0.33 \times 10^{11} \mathrm{NC}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(a) Inside the shell \(A\), electric field \(E_{\text {in }}=0\)

At the surface of shell \(A\),
\(
E_A=\frac{k Q_A}{r_A^2} \text { [a fixed positive value] }
\)
Between the shell \(A\) and \(B\), at a distance \(x\) from the common centre,
\(
E=\frac{k \cdot Q_A}{x^2} \quad[\text { as } x \text { increases, } E \text { decreases }]
\)
At the surface of shell \(B\), \(E_B=\frac{k\left(Q_A-Q_B\right)}{r_B^2} \quad\left[\right.\) a fixed negative value because \(\left.\left|Q_A\right|<\left|Q_B\right|\right]\)
Outside the both shell, at a distance \(x^{\prime}\) from the common centre,
\(
\left.E_{\text {out }}=\frac{k\left(Q_A-Q_B\right)}{x^{\prime 2}} \quad \begin{array}{l}
{\left[\text { as } x^{\prime} \text { increase negative value of } E_{\text {out }}\right.} \\
\text { decrease and it becomes zero at } x=\infty
\end{array}\right]
\)

Hint

\(
\begin{aligned}
&\text { (a) } \because T \sin \theta=F\\
&\text { Dividing the two equations, we get }\\
&\begin{aligned}
& =\frac{q}{4 \pi \varepsilon_0 x^2} \text { and } T \cos \theta=m g \\
\therefore \quad \tan \theta & =\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2 m g}
\end{aligned}
\end{aligned}
\)

\(
\frac{x}{2 l}=\frac{q^2}{4 \pi \varepsilon_0 x^2 m g} \quad\left(\because \tan \theta \approx \sin \theta=\frac{x}{2 l}\right)
\)
\(
\frac{x}{2 l} \propto \frac{q^2}{x^2} \text { or } q^2 \propto x^3
\)
\(
q \propto x^{\frac{3}{2}} \Rightarrow \frac{d q}{d t} \propto \frac{3}{2} x^{\frac{1}{2}} \frac{d x}{d t} \quad\left(\frac{d q}{d t}=\text { constant }\right)
\)
\(
v \propto x^{-\frac{1}{2}} \quad\left(\because v=\frac{d x}{d t}\right)
\)

Hint

(a) From figure, \(d l=R d \theta\),
Charge on \(d l, d q=\lambda R d \theta\)
Electric field at centre due to \(d l\) is
\(
d E=k \frac{\lambda R d \theta}{R^2} \quad\left(\because \lambda=\frac{q}{\pi R}\right)
\)

We need to consider only the component \(d E \cos \theta[latex], as the component [latex]d E \sin \theta\) will cancel out because of the field at \(C\) due to the symmetrical element \(d l^{\prime}\).
Total field at centre, \(|E|=2 \int_0^{\pi / 2} d E \cos \theta\)
\(
\begin{aligned}
& =\frac{2 k \lambda}{R} \int_0^{\pi / 2} \cos \theta d \theta \\
& =\frac{2 k \lambda}{R}=\frac{q}{2 \pi^2 \varepsilon_0 R^2}
\end{aligned}
\)

Hint

(b) At a point on the axis of uniformly charged disc at a distance \(x\) from the centre of the disc, the magnitude of the electric field is,
\(
E=\frac{\sigma}{2 \varepsilon_0}\left[1-\frac{x}{\sqrt{x^2+R^2}}\right]
\)
At centre,
\(
E_c=\frac{\sigma}{2 \varepsilon_0}
\)
Given that,
\(
\frac{E}{E_c}=\frac{1}{2}
\)
Then
\(
\begin{gathered}
1-\frac{x}{\sqrt{x^2+R^2}}=\frac{1}{2} \\
\frac{x}{\sqrt{x^2+R^2}}=\frac{1}{2}
\end{gathered}
\)
On squaring both sides, we get
Thus,
\(
x^2=\frac{x^2}{4}+\frac{R^2}{4}
\)

Hint

(a) The force acting on the electron \(=e E\).
Acceleration of the electron \(=\frac{e E}{m}\)
Here, \(\quad s=2 \times 10^{-2} \mathrm{~m}, u=0, v=\) ?
\(
\begin{array}{ll}
\therefore & v^2-u^2=2 a s \\
\Rightarrow & v^2=2\left(\frac{e}{m}\right) E \times 2 \times 10^{-2} \mathrm{~m}
\end{array}
\)
Also, \(\quad \frac{e}{m}=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}\)
\(
\begin{aligned}
\therefore \quad v^2 & =2 \times 1.76 \times 10^{11} \times 10^4 \times 2 \times 10^{-2} \\
& =7.04 \times 10^{13} \\
& =70.4 \times 10^{12}
\end{aligned}
\)
The velocity of the electron when it reaches plate \(B\),
\(
v \approx 0.85 \times 10^7 \mathrm{~m} / \mathrm{s}
\)

Hint

(a) In equilibrium,
\(
\begin{array}{ll}
& F_e=T \sin \theta \\
\text { and } & m g=T \cos \theta
\end{array}
\)

\(
\tan \theta=\frac{F_e}{m g}=\frac{q^2}{4 \pi \varepsilon_0 x^2 \times m g}
\)
Also, \(\tan \theta \approx \sin \theta=\frac{x / 2}{L}\)
Hence, \(\quad \frac{x}{2 L}=\frac{q^2}{4 \pi \varepsilon_0 x^2 \times m g}\)
\(x^3=\frac{2 q^2 L}{4 \pi \varepsilon_0 m g} \Rightarrow x=\left(\frac{q^2 L}{2 \pi \varepsilon_0 m g}\right)^{1 / 3}\)

Hint

(b) The three forces acting on each sphere are
(i) tension
(ii) weight
(iii) electrostatic force of repulsion

For sphere 1, in equilibrium, from the figure,
\(
\begin{array}{ll}
& T_1 \cos \theta_1=M_1 g \text { and } T_1 \sin \theta_1=F_1 \\
\therefore & \tan \theta_1=\frac{F_1}{M_1 g}
\end{array}
\)
For sphere 2, in equilibrium, from figure,
\(
\begin{aligned}
& T_2 \cos \theta_2=M_2 g \text { and } T_2 \sin \theta_2=F_2 \\
\therefore \quad & \tan \theta_2=\frac{F_2}{M_2 g}
\end{aligned}
\)
Force of repulsion between two charges are same.
\(
\therefore \quad F_1=F_2
\)
Here, \(\theta_1=\theta_2\) only, if \(\frac{F_1}{M_1 g}=\frac{F_2}{M_2 g} \Rightarrow M_1=M_2\)