Electrostatics – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.Download eSaral app for free study material and video tutorials.Simulator 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]

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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]

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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]

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Sol. (2)$\mathrm{q}=-100 \times 1.6 \times 10^{-19} \mathrm{C}$$\Delta \mathrm{V}=-14 \mathrm{volt}z\mathrm{W}=\mathrm{q} \Delta \mathrm{V}=2.24 \times 10^{-16} \mathrm{J} Q. \operatorname{Let} \mathrm{P}(\mathrm{r})=\frac{\mathrm{Q}}{\pi \mathrm{R}^{4}} \mathrm{r} be the charge density distribution for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r_{1} from the centre of the sphere, the magnitude of electric field is :-(1) \frac{\mathrm{Qr}_{1}^{2}}{4 \pi \epsilon_{0} \mathrm{R}^{4}}(2) \frac{\mathrm{Qr}_{1}^{2}}{3 \pi \epsilon_{0} \mathrm{R}^{4}}(3) 0(4) \frac{\mathrm{Q}}{4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}} [AIEEE – 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field \overrightarrow{\mathrm{E}} at the centre O is :-(1) \frac{\mathrm{q}}{2 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}( 2)\frac{\mathrm{q}}{4 \pi^{2} \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$$(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]

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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]

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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]$$=\frac{\rho_{0} r}{4 \in_{0}}\left[\frac{5}{3}-\frac{r}{R}\right] Q. Two identical charged spheres suspended from a common point by two massless string of length \ellare initially a distance d(d << \ell) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them :-(1) \mathrm{v} \propto \mathrm{x}^{1 / 2}(2) \mathrm{v} \propto \mathrm{x}(3) \mathrm{v} \propto \mathrm{x}^{-1 / 2}(4) \mathrm{v} \propto \mathrm{x}^{-1} [AIEEE – 2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)\tan \theta=\frac{\mathrm{F}}{\mathrm{M}_{9}} \quad(\text { since } \theta \text { small }) Q. The electrostatic potential inside a charged spherical ball is given by \phi=\mathrm{ar}^{2}+\mathrm{b}where r is the distance from the centre; a, b are constant. Then the charge density inside the ball is :-(1) -24 \pi \mathrm{a} \in_{0}$$(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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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 ofStatement-1.(2) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation ofstatement-1.(3) Statement-1 is true, Statement-2 is false(4) Statement-1 is false, Statement-2 is true [AIEEE – 2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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]

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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]

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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]

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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

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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]

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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]

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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]

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Sol. (1)

• December 29, 2021 at 3:42 pm

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• June 14, 2021 at 9:39 pm

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• July 21, 2021 at 7:46 am

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• May 11, 2021 at 1:49 pm

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• September 5, 2021 at 7:41 pm

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• May 4, 2021 at 7:21 am

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• May 2, 2021 at 5:22 pm

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• May 2, 2021 at 1:12 am

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• April 28, 2021 at 7:11 am

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• March 18, 2021 at 11:19 pm

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• February 10, 2021 at 1:18 pm

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• January 10, 2021 at 12:10 am

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• November 22, 2020 at 3:52 pm

APPRECIATION; THANKU SO MUCH, It was really helpful for me. :>
REQUEST: Plz…. upload chapter wise segregated questions from 2019 and 2020 jee papers.
BEFORE JEE MAINS 2021 JANUARY.

• April 20, 2021 at 9:23 pm

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Thanks me later😉

• October 21, 2020 at 3:09 pm

• October 11, 2020 at 10:29 am

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• November 16, 2020 at 12:10 am

• March 14, 2021 at 9:57 am

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• September 7, 2020 at 11:14 am

• September 3, 2020 at 12:35 pm

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• August 19, 2020 at 10:52 am

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• August 16, 2020 at 1:24 pm

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• August 9, 2020 at 10:53 am

• August 1, 2020 at 9:49 pm

Add some more questions on concentric spheres, spherical cell etc.

• July 12, 2020 at 5:24 pm

Thankyou bhut maja aaya but please or questions dal dijiye

• July 9, 2020 at 5:44 pm

• July 9, 2020 at 10:40 am

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• July 4, 2020 at 7:50 pm

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• June 30, 2020 at 12:53 pm

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• June 22, 2020 at 12:57 pm

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• 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

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• May 29, 2020 at 1:20 pm

Thank you sir
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• May 29, 2020 at 1:13 pm

• May 28, 2020 at 10:03 am

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• May 25, 2020 at 8:10 pm

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• May 25, 2020 at 7:30 pm

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• May 18, 2020 at 2:14 pm

Having kids

• May 3, 2020 at 8:48 pm

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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

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• April 30, 2020 at 10:32 am

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• April 29, 2020 at 2:44 pm