Electrostatics – JEE Main Previous Year Questions with Solutions

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Previous Years AIEEE/JEE Mains Questions

Q. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then $\frac{\mathrm{Q}}{\mathrm{q}}$ equals :-

(1) 1

$(2)-\frac{1}{\sqrt{2}}$

(3) $-2 \sqrt{2}$

(4) –1

[AIEEE – 2009]

Sol. (3)


Q. Statement–1 : For a charged particle moving from point P to point Q the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

Statement–2 : The net work done by a conservative force on an object moving along closed loop is zero.

(1) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement–1

(2) Statement–1 is false, Statement–2 is true

(3) Statement–1 is true, Statement–2 is false

(4) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement–

[AIEEE – 2009]

Sol. (4)

Electrostatics field is a conservative field work is independent of path.


Q. Two points P and Q are maintained at the potential of 10V and –4V, respectively. The work done in moving 100 electrons from P to Q is :-

(1) $-2.24 \times 10^{-16} \mathrm{J}$

(2) $2.24 \times 10^{-16} \mathrm{J}$

(3) $-9.60 \times 10^{-17} \mathrm{J}$

(4) $9.60 \times 10^{-17} \mathrm{J}$

[AIEEE – 2009]

Sol. (2)

$\mathrm{q}=-100 \times 1.6 \times 10^{-19} \mathrm{C}$

$\Delta \mathrm{V}=-14 \mathrm{volt}$z

$\mathrm{W}=\mathrm{q} \Delta \mathrm{V}=2.24 \times 10^{-16} \mathrm{J}$


Q. $\operatorname{Let} \mathrm{P}(\mathrm{r})=\frac{\mathrm{Q}}{\pi \mathrm{R}^{4}} \mathrm{r}$ be the charge density distribution for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance $r_{1}$ from the centre of the sphere, the magnitude of electric field is :-

(1) $\frac{\mathrm{Qr}_{1}^{2}}{4 \pi \epsilon_{0} \mathrm{R}^{4}}$

(2) $\frac{\mathrm{Qr}_{1}^{2}}{3 \pi \epsilon_{0} \mathrm{R}^{4}}$

(3) 0

(4) $\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}}$

[AIEEE – 2009]

Sol. (1)


Q. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field $\overrightarrow{\mathrm{E}}$ at the centre O is :-

(1) $\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

( 2)$\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

$(3)-\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

(4) $-\frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

[AIEEE – 2010]

Sol. (4)

$\overrightarrow{\mathrm{E}}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}}(-\hat{\mathrm{j}})=\frac{2}{\left(4 \pi \epsilon_{0} \mathrm{r}\right)} \frac{\mathrm{q}}{(\pi \mathrm{r})}(-\hat{\mathrm{j}})$


Q. Let there be a spherically symmetric charge distribution with charge density varying as

(r) = $\rho_{0}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$ upto r = R, and (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r < R) from the origion is given by :

( 1)$\frac{\rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(2) $\frac{4 \pi \rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(3) $\frac{\rho_{0} \mathrm{r}}{4 \varepsilon_{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(4) $\frac{4 \rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

[AIEEE – 2010]

Sol. (3)

Total charge

$Q=\int_{0}^{r} \rho d V=\int_{0}^{r} \rho_{0}\left(\frac{5}{4}-\frac{r}{R}\right) 4 \pi r^{2} d r$

$=4 \pi \rho_{0} \int_{0}^{r}\left(\frac{5 r^{2}}{4}-\frac{r^{3}}{R}\right) d r=4 \pi \rho_{0}\left[\frac{5 r^{3}}{12}-\frac{r^{4}}{4 R}\right]$

$\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}=\frac{1}{4 \pi \epsilon_{0} \mathrm{r}^{2}} 4 \pi \rho_{0}\left[\frac{5}{12} \mathrm{r}^{3}-\frac{\mathrm{r}^{4}}{4 \mathrm{R}}\right]$

$=\frac{\rho_{0} r}{4 \in_{0}}\left[\frac{5}{3}-\frac{r}{R}\right]$


Q. Two identical charged spheres suspended from a common point by two massless string of length $\ell$are initially a distance d(d << $\ell$) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them :-

(1) $\mathrm{v} \propto \mathrm{x}^{1 / 2}$

(2) $\mathrm{v} \propto \mathrm{x}$

(3) $\mathrm{v} \propto \mathrm{x}^{-1 / 2}$

(4) $\mathrm{v} \propto \mathrm{x}^{-1}$

[AIEEE – 2011]

Sol. (3)

$\tan \theta=\frac{\mathrm{F}}{\mathrm{M}_{9}} \quad(\text { since } \theta \text { small })$


Q. The electrostatic potential inside a charged spherical ball is given by $\phi=\mathrm{ar}^{2}+\mathrm{b}$where r is the distance from the centre; a, b are constant. Then the charge density inside the ball is :-

(1) $-24 \pi \mathrm{a} \in_{0}$

$(2)-6 \mathrm{a} \in_{0}$

(3) $-24 \pi \mathrm{a} \in_{0} \mathrm{r}$

$(4)-6 \mathrm{a} \in_{0} \mathrm{r}$

[AIEEE – 2011]

Sol. (2)


Q. Two positive charges of magnitude ‘q’ are placed at the ends of a side (side 1) of a square of side ‘2a’. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is :-

(1) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1-\frac{1}{\sqrt{5}}\right)$

(2) zero

(3) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1+\frac{1}{\sqrt{5}}\right)$

(4) $\frac{1}{4 \pi \epsilon_{0}} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1-\frac{2}{\sqrt{5}}\right)$

[AIEEE – 2011]

Sol. (1)


Q. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

An insulating solid sphere of radius R has a uniformaly positive charge density . As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphre and also at a point out side the sphere. The electric potential at infinity is zero.

Statement-1: When a charge ‘q’ is taken from the centre to the surface of the sphere, its potential energy changes by $\frac{\mathrm{q} \rho}{3 \epsilon_{0}}$

Statement-2 : The electric field at a distance r (r < R) from the centre of the sphere is $\frac{\rho \mathrm{r}}{3 \epsilon_{0}}$

(1) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of

Statement-1.

(2) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of

statement-1.

(3) Statement-1 is true, Statement-2 is false

(4) Statement-1 is false, Statement-2 is true

[AIEEE – 2012]

Sol. (4)


Q. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be :-

[AIEEE – 2012]

Sol. (4)

For uniformly charged sphere

$\mathrm{E}=\frac{\mathrm{Kqr}}{\mathrm{R}^{3}}(\mathrm{r}<\mathrm{R})$

$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{R}^{2}} \quad(\mathrm{r}=\mathrm{R})$

$\mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{r}^{2}} \quad(\mathrm{r}>\mathrm{R})$


Q. Let $\left[\in_{0}\right]$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = Length,

T = Time and A = electric current, then :

(1) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^{2} \mathrm{A}\right]$

(2) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^{4} \mathrm{A}^{2}\right]$

(3) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-1} \mathrm{A}^{-2}\right]$

(4) $\left[\in_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-1} \mathrm{A}\right]$

[AIEEE – 2013]

Sol. (2)


Q. Two charges, each equal to q, are kept at x = –a and x = a on the x-axis. A particle of mass m and charge $\mathrm{q}_{0}=\frac{\mathrm{q}}{2}$ is placed at the origin. If charge $q_{0}$ is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to

(1) y           (2) –y          (3) $\frac{1}{y}$ $          (4)-\frac{1}{\mathrm{y}}$

[AIEEE – 2013]

Sol. (1)


Q. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :-

(1) $\frac{\mathrm{Q}}{8 \pi \epsilon_{0} \mathrm{L}}$

(2) $\frac{3 \mathrm{Q}}{4 \pi \in_{0} \mathrm{L}}$

(3) $\frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{L} \ln 2}$

(4) $\frac{\mathrm{Q} \ln 2}{4 \pi \epsilon_{0} \mathrm{L}}$

[JEE-Main-2013]

Sol. (4)


Q. Assume that an electric field $\overrightarrow{\mathrm{E}}=30 \mathrm{x}^{2} \hat{\mathrm{i}}$ exists in space. Then the potential difference $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{O}}$, where $\mathrm{V}_{\mathrm{O}}$ is the potential at the origin and $\mathrm{V}_{\mathrm{A}}$ the potential at x = 2 m is :-

(1)–80 J

(2) 80 J

(3) 120 J

(4) –120 J

Sol. (1)

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}=-10[8-0]=-80 \mathrm{V}$


Q. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density $\rho=\frac{\mathrm{A}}{\mathrm{r}}$, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :-

(1) $\frac{2 Q}{\pi a^{2}}$

(2) $\frac{\mathrm{Q}}{2 \pi \mathrm{a}^{2}}$

(3) $\frac{\mathrm{Q}}{2 \pi\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right)}$

(4) $\frac{2 \mathrm{Q}}{\pi\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}$

[JEE-Main-2016]

Sol. (2)

Gaussian surface at distance r from center


Q. An electric dipole has a fixed dipole moment $\overrightarrow{\mathrm{p}}$, which makes angle  with respect to x-axis. When subjected to an electric field $\overrightarrow{\mathrm{E}}_{1}=\mathrm{E} \hat{\mathrm{i}}$, it experiences a torque $\overrightarrow{\mathrm{T}}_{1}=\tau \hat{\mathrm{k}}$. When subjected to another electric field $\overrightarrow{\mathrm{E}}_{2}=\sqrt{3} \mathrm{E}_{1} \hat{\mathrm{j}}$ it experiences torque . The angle $\theta$ is :

(1) $60^{\circ}$ (2) $90^{\circ}$ (3) $30^{\circ}$ (4) $45^{\circ}$

[JEE-Main-2017]

Sol. (1)

So from $\quad \vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}$


Q. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities $+\sigma,-\sigma$ and $+\sigma$ respectively. The potential of shell B is :-

[JEE-Main-2018]

Sol. (1)


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Comments
  • July 9, 2020 at 5:44 pm

    Thankyou, it was very helpful

  • July 9, 2020 at 10:40 am

    Thank you
    Really esaral is the best site ……

  • July 4, 2020 at 7:50 pm

    U should give nice explanation and useful questions

  • June 30, 2020 at 12:53 pm

    Gjici

  • June 22, 2020 at 12:57 pm

    thank you but add some more questions of recent years

  • June 19, 2020 at 6:27 pm

    very nice…..

  • June 13, 2020 at 2:24 pm

    Thank you so much but add some questions on Gauss law

  • June 8, 2020 at 1:33 am

    Thank u very much and please add some 2019and 2020 questions also

  • June 2, 2020 at 11:07 am

    great work

  • May 29, 2020 at 1:20 pm

    Thank you sir
    This is an easier way for JEE aspirants

  • May 29, 2020 at 1:13 pm

    Thank you very much……but please add some more questions

  • May 28, 2020 at 10:03 am

    We need more

  • May 25, 2020 at 8:10 pm

    Awesome work.E Saral will soon be one of the best online platform for competitive exams

  • May 25, 2020 at 7:30 pm

    Thanks sir

  • May 18, 2020 at 2:14 pm

    Having kids

  • May 3, 2020 at 8:48 pm

    Thank you so much sir it’s so much useful
    But I think that AIEEE 2010 solution was wrong and the process would be by taking Gaussian surface

  • May 2, 2020 at 1:08 pm

    Thanks sir

  • April 30, 2020 at 10:32 am

    Thanks for such a great helping

  • April 29, 2020 at 2:44 pm

    sir,please upload 2019 and 2020 questions also.

  • April 26, 2020 at 3:15 pm

    Thanks for providing the questions of previous years.

  • April 26, 2020 at 1:32 pm

    Thank you very much! Wish you succeed in your work and hope that work will be to provide the stuff like this.

  • March 26, 2020 at 8:58 am

    Very nice