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**Previous Years JEE Advanced Questions**

$(\mathrm{A}) \frac{-2 C}{\varepsilon_{o}}$

$(\mathrm{B}) \frac{2 C}{\varepsilon_{o}}$

$(\mathrm{C}) \frac{10 C}{\varepsilon_{o}}$

$(\mathrm{D}) \frac{12 C}{\varepsilon_{o}}$

** [IIT-JEE 2009]**

**Sol.**(A)

(A) 1 : 2 : 3 (B) 1 : 3 : 5 (C) 1 : 4 : 9 (D) 1 : 8 : 18

**[IIT-JEE 2009]**

**Sol.**(B)

(A) The angular momentum of the charge –q is constant

(B) The linear momentum of the charge –q is constant

(C) The angular velocity of the charge –q is constant

(D) The linear speed of the charge –q is constant

** [IIT-JEE 2009]**

**Sol.**(A)

Since torque about central charge is zero angular momentum is conserved.

*R*has a charge

*Q*distributed in its volume with a charge density =

*$K r^{\mathrm{a}}$*, where

*K*and a are constants and

*r*is the distance from its centre. If the electric field at

*$r=\frac{R}{2}$*is $\frac{1}{8}$ times that at

*r*=

*R*, find the value of

*a*.

** [IIT-JEE 2009]**

**Sol.**2

*R*carries uniform surface charge density of $\sigma$ per unit area. It is made of two hemispherical shells, held together by pressing them with force

*F*(see figure).

*F*is proportional to –

(A) $\frac{1}{\varepsilon_{0}} \sigma^{2} R^{2}$

(B) $\frac{1}{\varepsilon_{0}} \sigma^{2} R$

(C) $\frac{1}{\varepsilon_{0}} \frac{\sigma^{2}}{R}$

$(\mathrm{D}) \frac{1}{\varepsilon_{0}} \frac{\sigma^{2}}{R^{2}}$

**[IIT-JEE 2010]**

**Sol.**(A)

Electrostatic pressure $\mathrm{P}=\frac{\sigma^{2}}{2 \epsilon_{0}} \Rightarrow \mathrm{F}=\frac{\sigma^{2}}{2 \epsilon_{0}} \cdot \pi \mathrm{R}^{2} \Rightarrow \mathrm{F} \propto \frac{\sigma^{2} \mathrm{R}^{2}}{\epsilon_{0}}$

*q*is –

(A) $1.6 \times 10^{-19} \mathrm{C}$

(B) $3.2 \times 10^{-19} \mathrm{C}$

(C) $4.8 \times 10^{-19} \mathrm{C}$

(D) $8.0 \times 10^{-19} \mathrm{C}$

**[IIT-JEE 2010]**

**Sol.**(D)

$\mathrm{qE}=\mathrm{mg} \ldots(\mathrm{i}) ; \mathrm{mg}=6 \mathrm{phrV} \ldots(\mathrm{ii})$ and $\mathrm{m}=\frac{4}{3} \pi \mathrm{r}^{3} \mathrm{gp} \ldots(\mathrm{iii})$

On solving above equations we get $q=8.0 \times 10^{-19} \mathrm{C}$

*$Q_{1}$ and $Q_{2}$*fixed at two different points on the

*x*-axis are shown in the figure. These lines suggest that –

(A) $\left|Q_{1}\right|>\left|Q_{2}\right|$

(B) $\left|Q_{1}\right|<\left|Q_{2}\right|$

(C) at a finite distance to the left of *$Q_{1}$* the electric field is zero

(D) at a finite distance to the right of *$Q_{2}$* the electric field is zero

**[IIT-JEE 2010]**

**Sol.**(A,D)

Number of lines of force exiting from $\mathrm{Q}_{1}$ is greater than that of enetering in $\mathrm{Q}_{2} \Rightarrow\left|\mathrm{Q}_{1}\right|>\left|\mathrm{Q}_{2}\right|$

$\frac{\mathrm{KQ}_{1}}{(\ell+\mathrm{x})^{2}}=\frac{\mathrm{KQ}_{2}}{\left(\mathrm{x}^{2}\right)} \Rightarrow \frac{\mathrm{x}}{(\ell+\mathrm{x})}=\sqrt{\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}}$ because $\mathrm{x}>0$ so there will be point in right of $\mathrm{Q}_{2}$

(A) $2 E_{0} a^{2}$

(B) $\sqrt{2} E_{0} a^{2}$

$(\mathrm{C}) E_{0} a^{2}$

(D) $\frac{E_{0} a^{2}}{\sqrt{2}}$

**[IIT-JEE 2011]**

**Sol.**(C)

$\phi=\int \bar{E} \cdot \overrightarrow{d S}=$ Ex projected area perpendiuclar to E $(\mathrm{x}-\mathrm{axis})=\mathrm{E} \times \mathrm{a}^{2}$

(A) $E_{A}^{\text {inside }}=0$

(B) $Q_{A}>Q_{B}$

(C) $\frac{\sigma_{A}}{\sigma_{B}}=\frac{R_{B}}{R_{A}}$

(D) $E_{A}^{\text {on sufface }}<E_{B}^{\text {on sufface }}$

**[IIT-JEE 2011]**

**Sol.**(A,B,C,D)

$\mathrm{E}_{A}^{\text {inside }}=0$ (because of electrostatic condition) So, A option is true.

$\Rightarrow \mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{B}} \Rightarrow \frac{k Q_{A}}{R_{A}}=\frac{k Q_{B}}{R_{B}} \Rightarrow \frac{Q_{A}}{Q_{B}}=\frac{R_{A}}{R_{B}} \Rightarrow \mathrm{R}_{\mathrm{B}}<\mathrm{R}_{\mathrm{A}} \mathrm{So}, \mathrm{Q}_{\mathrm{B}}<\mathrm{Q}_{\mathrm{A}},$ so $\mathrm{B}$ is true

$\Rightarrow \frac{\sigma_{A} 4 \pi R_{A}^{2}}{\sigma_{B} 4 \pi R_{B}^{2}}=\frac{R_{A}}{R_{B}} \Rightarrow \frac{\sigma_{A}}{\sigma_{B}}=\frac{R_{B}}{R_{A}},$ So $\mathrm{C}$ is true

$\mathrm{E}_{\text {nexsurixe }}=\sigma \times \frac{1}{R} \cdot \mathrm{So}, \mathrm{D}$ is also true

*Q*on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown, then the SHM of the block will be –

(A) of the same frequency and with shifted mean position

(B) of the same frequency and with the same mean position

(C) of changed frequency and with shifted mean position

(D) of changed frequency and with the same mean position

**[IIT-JEE 2011]**

**Sol.**(A)

Time period of spring block system depends on spring constant and mass of block.

On applying electric field only the equilibrium position gets shifted.

(A) If the electric field due to a point charge varies as *$r^{-2.5}$ instead of $r^{-2}$*, then the Gauss law will still be valid.

(B) The Gauss law can be used to calculate the field distribution around an electric dipole.

(C) If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.

(D) The work done by the external force in moving a unit positive charge from point *A* at potential *$V_{A}$ to point $\mathrm{B}$ at potential $V_{B}$ is $\left(V_{B}-V_{A}\right)$*

**[IIT-JEE 2011]**

**Sol.**(C,D)

The field distribution for a dipole can not be calculated by using Gauss law only, therefore (C,D)

**[IIT-JEE 2011]**

**Sol.**3

**Paragraph for Question Nos. 13 and 14**

A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let ‘N’ be the number density of free electrons, each of mass ‘m’. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begins to oscillate about the positive ions with a natural angular frequency ‘$\omega_{\rho}$’, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency $\omega$, where a part of the energy is absorbed and a part of it is reflected. As $\omega$ approaches $\omega_{p}$, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals

$(\mathrm{A}) \sqrt{\frac{N e}{m \varepsilon_{0}}}$

(B) $\sqrt{\frac{m \varepsilon_{0}}{N e}}$

(C) $\sqrt{\frac{N e^{2}}{m \varepsilon_{0}}}$

$(\mathrm{D}) \sqrt{\frac{m \varepsilon_{0}}{N e^{2}}}$

**[IIT-JEE 2011]**

**Sol.**(C)

Dimension of $\omega_{P}$ is $T^{-1}$

$[\sqrt{\frac{N e^{2}}{n E_{0}}}]=T^{-1}$

$2 \pi f=\omega P=\sqrt{\frac{N e^{2}}{n E_{0}}}$

(A) 800 nm (B) 600 nm (C) 300 nm (D) 200 nm

**[IIT-JEE 2011]**

**Sol.**(B)

$\omega=\sqrt{\frac{N e^{2}}{m E_{0}}} 2 \pi f$

$\lambda=\frac{C}{f}$

**[IIT-JEE 2012]**

**Sol.**6

Electric field due to cylinder $=\frac{2 K \lambda}{2 R}$

Electric field due to sphere $=\frac{K Q}{(2 R)^{2}}\left[\text { where } K=\frac{1}{4 \pi \epsilon_{0}}\right]$

Electric field due to sphere $=\frac{K Q}{(2 R)^{2}}\left[\text { where } K=\frac{1}{4 \pi \epsilon_{0}}\right]$

$E=\frac{2 K(\lambda)}{2 R}-\frac{K(Q)}{(2 R)^{2}}=\frac{K\left(\rho \pi R^{2}\right)}{R}-\frac{K\left(\rho \times \frac{4 \pi}{3}\left(\frac{R^{3}}{8}\right)\right)}{4 R^{2}}$

$=K(\rho \pi R)-\frac{K(\rho)(\pi R)}{24}=\frac{23 \rho \pi R}{4 \pi \varepsilon_{0} \times 24}=\frac{23 \rho R}{16 \times 6 \varepsilon_{0}} \Rightarrow k=6$

(A) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane x = – a/2

(B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = –a/2.

(C) The net electric flux crossing the entire region is $\frac{q}{\varepsilon_{0}}$

(D) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane x = +a/2.

**[IIT-JEE 2012]**

**Sol.**(A,C,D)

For (A): Position of charges is symmetrical with respect to planes at x = + a/2 and x = – a/2

For (C) : Net charge enclosed by cube is q.

For (D) : Charges are symmetrically placed iwth given planes.

(A) $1 \times 10^{-5} \mathrm{V}$

(B) $1 \times 10^{-7} \mathrm{V}$

(C) $1 \times 10^{-9} \mathrm{V}$

(D) $1 \times 10^{-10} \mathrm{V}$

**[IIT-JEE 2012]**

**Sol.**(C)

As proton moves at $45^{\circ}$ in vertical plane it implies magnitude of electric force and gravitational force is same.

for $45^{\circ}$ direction $|\mathrm{mg}|=|\mathrm{qE}|$

$\frac{m g}{q}=E=\frac{V}{d}$

$V=\frac{m g d}{q}=\frac{\left(1.6 \times 10^{-27} \mathrm{kg}\right)\left(1 \times 10^{-2} \mathrm{m}\right)}{\left(1.6 \times 10^{-19} \mathrm{C}\right)}=1 \times 10^{-9} \mathrm{volt}$

**[IIT-JEE 2012]**

**Sol.**(D)

Electric field inside a uniformly charged spherical shell :

(A) The electric field at O is 6 K along OD.

(B) The potential at O is zero.

(C) The potential at all points on the line PR is same.

(D) The potential at all points on the line ST is same.

**[IIT-JEE 2012]**

**Sol.**(A,B,C)

Resultant is 6 K along OD. For every point of POR there are two charges of opposite sign and equal magnitude at equal distance.

(A) $-4$

$(\mathrm{B})-\frac{32}{25}$

(C) $\frac{32}{25}$

(D) 4

**[JEE-Advance-2013]**

**Sol.**(B,D)

$\mathrm{q}_{1}=\rho_{1} \frac{4}{3} \pi \mathrm{R}^{3} ; \mathrm{q}_{2}=\rho_{2} \frac{4}{3} \pi(2 \mathrm{R})^{3}$

if $\mathrm{E}_{\mathrm{net}}=0$ at point A then

$\frac{\mathrm{kq}_{1}}{(2 \mathrm{R})^{2}}+\frac{\mathrm{kq}_{2}}{(5 \mathrm{R})^{2}}=0 \Rightarrow \frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}=-\frac{4}{25}$

$\Rightarrow \frac{\rho_{1}}{\rho_{2}}=-\frac{32}{25}$

if $\mathrm{E}_{\mathrm{net}}=0$ at point $\mathrm{B}$ then

$\frac{\mathrm{kq}_{1}}{(2 \mathrm{R})^{2}}=\frac{\mathrm{k} \mathrm{q}_{2} \mathrm{R}}{(2 \mathrm{R})^{3}} \Rightarrow \frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}=\frac{1}{2} \Rightarrow \frac{\rho_{1}}{\rho_{2}}=4$

(A) the electrostatic field is zero

(B) the electrostatic potential is constant

(C) the electrostatic field is constant in magnitude

(D) the electrostatic field has same direction

**[JEE-Advance-2013]**

**Sol.**(C,D)

Electric field at point $P=\frac{\rho(\overline{A P})}{3 \epsilon_{0}}+\frac{-\rho(\overline{B P})}{3 \epsilon_{0}}$

$=\frac{\rho(\overline{\mathrm{AP}}-\overline{\mathrm{BP}})}{3 \epsilon_{0}}=\frac{\rho(\overline{\mathrm{AB}})}{3 \epsilon_{0}}$

= constant in both magnitude and direction at every point of overlap region.

Also E is non zero so potential is not same at all points.

(A) $\mathrm{Q}=4 \sigma \pi \mathrm{r}_{0}^{2}$

(B) $\mathrm{r}_{0}=\frac{\lambda}{2 \pi \sigma}$

(C) $\mathrm{E}_{1}\left(\mathrm{r}_{0} / 2\right)=2 \mathrm{E}_{2}\left(\mathrm{r}_{0} / 2\right)$

(D) $\mathrm{E}_{2}\left(\mathrm{r}_{0} / 2\right)=4 \mathrm{E}_{3}\left(\mathrm{r}_{0} / 2\right)$

**[JEE-Advance-2014]**

**Sol.**(C)

$(\mathrm{A}) \mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$

$(\mathrm{B}) \mathrm{E}_{3}>\mathrm{E}_{1}>\mathrm{E}_{2}$

$(\mathrm{C}) \mathrm{E}_{2}>\mathrm{E}_{1}>\mathrm{E}_{3}$

$(\mathrm{D}) \mathrm{E}_{3}>\mathrm{E}_{2}>\mathrm{E}_{1}$

**[JEE-Advance-2014]**

**Sol.**(C)

(A) P-3, Q-1, R-4, S-2

(B) P-4, Q-2, R-3, S-1

(C) P-3, Q-1, R-2, S-4

(D) P-4, Q-2, R-1, S-3

**[JEE-Advance-2014]**

**Sol.**(A)

When all are positive net field is along +y axis

$\mathrm{P} \rightarrow 3$

when $\mathrm{Q}_{1}, \mathrm{Q}_{2}$ positive $\mathrm{Q}_{3}, \mathrm{Q}_{4}$ negative

$\mathrm{Q} \rightarrow 1$

When $\mathrm{Q}_{1}, \mathrm{Q}_{4}$ positive and $\mathrm{Q}_{2}, \mathrm{Q}_{3}$ negative

$\mathrm{R} \rightarrow 4$

**[JEE-Advance-2015]**

**Sol.**6

Every point on circumference of flat surface is at equal distance from point charge

Hence circumference is equipotential.

Flux passing through curved surface = – flux passing through flat surface.

Note : Flux through surface can be calculated using concept of solid angle.

(A) Both charges execute simple harmonic motion

(B) Both charges will continue moving in the direction of their displacement

(C) Charge +q executes simple harmonic motion while charge –q continues moving in the direction of its displacement.

(D) Charge –q executes simple harmonic motion while charge +q continues moving in the direction of its displacement.

**[JEE-Advance-2015]**

**Sol.**(C)

Field due to straight wire is perpendicular to the wire & radially outward. Hence $\mathrm{E}_{\mathrm{z}}=0$ Length, PQ = 2R sin 60 = $\sqrt{3} \mathrm{R}$ According to Gauss’s law

total flux $=\left[\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\frac{q_{\mathrm{in}}}{\epsilon_{0}}=\frac{\lambda \sqrt{3} \mathrm{R}}{\epsilon_{0}}\right.$

(A) $\overrightarrow{\mathrm{E}}$ is uniform, its magnitude is independent of $\mathrm{R}_{2}$ but its direction depends on $\overrightarrow{\mathrm{r}}$

(B) $\overrightarrow{\mathrm{E}}$ is uniform, its magnitude depends on $\mathrm{R}_{2}$ and its direction depends on $\overrightarrow{\mathrm{r}}$

(C) $\overrightarrow{\mathrm{E}}$ is uniform, its magnitude is independent of a but its direction depends on $\vec{a}$

(D) $\overrightarrow{\mathrm{E}}$ is uniform and both its magnitude and direction depend on $\vec{a}$

**[JEE-Advance-2015]**

**Sol.**(D)

$(\mathrm{A}) \ell=\sqrt{\left(\frac{\mathrm{nq}^{2}}{\varepsilon \mathrm{k}_{\mathrm{B}} \mathrm{T}}\right)}$

$(\mathrm{B})^{\ell}=\sqrt{\left(\frac{\varepsilon \mathrm{k}_{\mathrm{B}} \mathrm{T}}{\mathrm{nq}^{2}}\right)}$

(C) $\ell=\sqrt{\left(\frac{\mathrm{q}^{2}}{\mathrm{n}^{2 / 3} \mathrm{k}_{\mathrm{B}} \mathrm{T}}\right)}$

(D) $\ell=\sqrt{\left(\frac{\mathrm{q}^{2}}{\varepsilon \mathrm{n}^{1 / 3} \mathrm{k}_{\mathrm{B}} \mathrm{T}}\right)}$

**[JEE-Advance-2016]**

**Sol.**(B,D)

In first case if q is shifted towards one of the wires, its repulsive force will increase which will bring the charge back.

In second case, in same type of displacement attractive force due to wire increases which will make it move further towards wire.

$\therefore$ (C)

(A) The circumference of the flat surface is an equipotential

(B) The electric flux passing through the curved surface of the hemisphere is $-\frac{\mathrm{Q}}{2 \varepsilon_{0}}\left(1-\frac{1}{\sqrt{2}}\right)$

(C) Total flux through the curved and the flat surfaces is $\frac{\mathrm{Q}}{\varepsilon_{0}}$

(D) The component of the electric field normal to the flat surface is constant over the surface.

**[JEE-Advance-2017]**

**Sol.**(A,B)

Considering any point A insde the cavity electric field can be calculated at A by using super position priciple.

$\overrightarrow{\mathrm{E}}_{\mathrm{A}}=\left(\overrightarrow{\mathrm{E}}_{\mathrm{A}}\right)_{\mathrm{Due} \text { to sphere without cavity }}-\left(\overrightarrow{\mathrm{E}}_{\mathrm{A}}\right)_{\text {Due to sphere of radius } \mathrm{R}_{2} \& \text { centre at } \mathrm{P}}$

$\overrightarrow{\mathrm{E}}_{\mathrm{A}}=\frac{\rho}{3 \epsilon_{0}} \mathrm{O} \overrightarrow{\mathrm{A}}=\frac{\rho}{3 \epsilon_{0}}(\mathrm{P} \overrightarrow{\mathrm{A}})$

But $\mathrm{O} \overrightarrow{\mathrm{P}}+\mathrm{P} \overrightarrow{\mathrm{A}}=\mathrm{O} \overrightarrow{\mathrm{P}}$

$\therefore \mathrm{O} \overrightarrow{\mathrm{A}}-\mathrm{P} \overrightarrow{\mathrm{A}}=\mathrm{O} \overrightarrow{\mathrm{P}}$

$\therefore \overrightarrow{\mathrm{E}}_{\mathrm{A}}=\frac{\rho}{3 \epsilon_{0}}(\mathrm{O} \overrightarrow{\mathrm{P}})=\frac{\rho}{3 \epsilon_{0}}(\overrightarrow{\mathrm{a}})$

So E.F. is independent of location of A inside the cavity.

Ans. (D) is correct

(A) The electric flux through the shell is $\sqrt{3} \mathrm{R} \lambda / \varepsilon_{0}$

(B) The z-component of the electric field is zero at all the points on the surface of the shell

(C) The electric flux through the shell is $\sqrt{2} \mathrm{R} \lambda / \varepsilon_{0}$

(D) The electric field is normal to the surface of the shell at all points

**[JEE-Advance-2018]**

**Sol.**(A,B)

We know,

So, dimension

**[JEE-Advance-2018]**

**Sol.**2

$\mathrm{n}=10^{-3} \mathrm{kg} \mathrm{q}=1 \mathrm{C} \mathrm{t}=0$

Force on particle will be

$\mathrm{F}=\mathrm{q} \mathrm{E}=\mathrm{q} \mathrm{E}_{0} \sin \omega \mathrm{t}$

at $\mathrm{v}_{\max }, \mathrm{a}, \mathrm{F}=0$

$\mathrm{qE}_{0} \sin \omega \mathrm{t}=0$

$\mathrm{F}=\mathrm{qE}_{0} \sin \omega \mathrm{t}$

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\mathrm{q} \frac{\mathrm{E}_{0}}{\mathrm{m}} \sin \omega \mathrm{t}$

$\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\int_{0}^{\pi / \omega} \frac{\mathrm{q} \mathrm{E}_{0}}{\mathrm{m}} \sin \omega \mathrm{t} \mathrm{dt}$

$\mathrm{v}-0=\frac{\mathrm{q} \mathrm{E}_{0}}{\mathrm{m} \omega}[-\cos \omega \mathrm{t}]_{0}^{\pi / \omega}$

$\mathrm{v}-0=\frac{\mathrm{qE}_{0}}{\mathrm{m} \omega}[(-\cos \pi)-(-\cos 0)]$

$\mathrm{v}=\frac{1 \times 1}{10^{-3} 10^{3}} \times 2=2 \mathrm{m} / \mathrm{s}$

Ans. 2. 00

**[JEE-Advance-2018]**

**Sol.**(B)

(i) $\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{d}^{2}} \Rightarrow \mathrm{E} \propto \frac{1}{\mathrm{d}^{2}}$

(ii) Dipole

$\mathrm{E}=\frac{2 \mathrm{kp}}{\mathrm{d}^{3}} \sqrt{1+3 \cos ^{2} \theta}$

$\mathrm{E} \propto \frac{1}{\mathrm{d}^{3}}$ for dipole

(iii) For line charge

$\mathrm{E}=\frac{2 \mathrm{k} \lambda}{\mathrm{d}}$

$\mathrm{E} \propto \frac{1}{\mathrm{d}}$

(iv) $\mathrm{E}=\frac{2 \mathrm{K} \lambda}{\mathrm{d}-\ell}-\frac{2 \mathrm{K} \lambda}{\mathrm{d}+\ell}$

$=2 \mathrm{K} \lambda\left[\frac{\mathrm{d}+\ell-\mathrm{d}+\ell}{\mathrm{d}^{2}-\ell^{2}}\right]$

$\mathrm{E}=\frac{2 \mathrm{K} \lambda(2 \ell)}{\mathrm{d}^{2}\left[1-\frac{\ell^{2}}{\mathrm{d}^{2}}\right]}$

$\mathrm{E} \propto \frac{1}{\mathrm{d}^{2}}$

(v) Electric field due to sheet

$\epsilon=\frac{\sigma}{2 \epsilon_{0}}$

$\in=\mathrm{v}$ is independent of $\mathrm{r}$

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