Electrostatics Previous year Questions with Solutions
Frequently Asked Questions
Practising PYQs helps recognize the types of questions that are typically asked, the complexity level, and the recurring themes within Electrostatics. Regular exposure to previous exam questions improves your analytical abilities and problem-solving speed, which are crucial for performing well under the time constraints of the JEE Main exam.
Aspirants can pinpoint the subtopics within Electrostatics frequently asked, such as Coulomb's Law, Electric Fields,Potential and Capacitors, allowing you to prioritize these areas during your preparation.
Electrostatics is a fundamental component of the JEE Main Physics syllabus. Historically, this chapter accounts for approximately 3.3% of the total marks in the Physics section. Candidates can expect around 2 to 3 questions from Electrostatics in each exam session.
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Electrostatics Previous Year Questions with Solutions
Electrostatics is the branch of Physics that deals with the study of stationary electric charges and their interactions. It involves understanding the forces, fields, and potential associated with electric charges at rest.
Previous year questions with solutions on Electrostatics provide a clear understanding of fundamental concepts like Coulomb's law, electric fields, potential, and capacitance, which are crucial for solving problems in the JEE Main exam. By mastering these topics, you can tackle various question types effectively, ensuring a strong grasp of the core principles that often appear in the exam. This will enhance problem-solving skills and improve accuracy in the test. In the JEE Main Physics exam, you can generally expect 2 to 3 questions from the Electrostatics chapter.
JEE Main Previous Year Solved Questions on Electrostatics
JEE Adv Previous Year Solved Questions on Electrostatics
Related Videos:
1.0Key Concepts to Remember
Definitions
1. Coulomb’s Law- The electrostatic force between two point charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. This force acts along the line connecting the two charges at all times.
F=Kr2q1q2k=4πε01=9×109Nm2C−2= Electrostatic constant or Coulomb’s Constant
2. Electric Field Intensity
The space around a charge or charge distribution, where another charge experiences an electric force, is called an electric field.
Mathematically, E=q0F,SI unit: N/C or V/m
3. Gauss Law- According to this law the total electric flux () through any closed surface (S) in free space is equal to 10 times the total electric charge (q) enclosed by the surface.
ϕ=∮E⋅dS=ε0qenclosed
4. Electric potential
It is defined as the work done by an external agent in taking a unit positive charge from a reference point to that point without changing its Kinetic Energy.
Vp=q(won−woff)
5. Electrical Capacitance(C)
It shows the capacity of a conductor to store electric charge.
Capacitance of parallel plate when dielectric is partially filled
C=(d−t)+ktε0A
Force between plates of parallel plate capacitor
F=2Aε0Q2=2d2dA−CV2
Pressure on each plate of a capacitor
P=2ε0σ2
2.0JEE Main Past Year Questions with Solutions on Electrostatics
Q-1. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle with each other. When suspended in water the angle remains the same. If density of the material of the sphere is 1.5 g/cc , the dielectric constant of water will be_____
(Take density of water=1g/cc )
Solution:
Case-1 When suspended in water
Tcosθ1=mgTsinθ1=FDividing both equations, we get,tanθ1=mgF...........(1)
Case-2 When suspense in water,value of electrostatic force may changes,
F′=KFF′=Tsinθ1.......(2)mg−U=Tcosθ1(whereU=Upthrust)σVg−ρVg=Tcosθ1(σ−ρ)Vg=Tcosθ1.˙........(3)Dividing (2) by (3):tanθ1=(σ−ρ)VgF′Dividing (1) by (4): 1=mgF×F(σ−ρ)Vg=σVgF×F(σ−ρ)Vg⇒1=K×σ(σ−ρ)×1.51.5−1=K×1.50.5=K×31⇒K=3K=3
Q-2. Two charges of 5Q and -2Q are situated at the points 3a,0 and -5a,0 respectively. The electric flux through a sphere of radius '4a 'having center at origin is :
(1)ε02Q(2)ε05Q(3)ε07Q(4)ε03Q
Solution: (2) ε05Q
5Q charge is inside the spherical region
ϕnet=ε0qenclosed=ε05Q
Q-3. An electric field is given by (6i^+5j^+3k^)N/C . The electric flux through a surface area 30i^m2 lying in YZ- plane (in SI unit) is
Q-4. Two identical charged spheres are suspended by a string of equal lengths. The string makes an angle of 30° with each other. When suspended in a liquid of density 0.8 gcm−3, the angle remains the same. If density of material of the sphere is 1.6 gcm−3 , the dielectric constant of the liquid is_____
Q-5. A particle of charge '-q' and mass 'm' moves in a circle a of radius 'r' around an infinitely long line charge of linear charge density '+'.Then time period will be given as Consider k as Coulomb's Constant
Q-6. Two charges q and 3q are separated by a distance 'r' in the air.At a distance x from charge q the resultant electric field is zero.The value of x is
Q-7. Force between two point charges q1andq2 placed in vacuum at 'r' cm apart is F. Force between them when placed in a medium having dielectric K=5 at 'r5'cm apart will be
(1)F=325(2)5F(3)5F(4)25F
Solution:(2) 5F
In air, F=4πε01⋅r2q1q2
In medium, F=4π(kε0)1⋅r2q1q2
=4πε0r2q1q2⋅k1=4πε025⋅r2q1q2⇒5F
Q-8. The distance between charges +q and -q is 2l and between +2q and -2q is 4l. The electrostatic potential at point P at a distance r from centre O is −α(r2ql)×109Vwhere the value of α is
Solution:
Due to q:Vq=rkqcosθ=rkq(2θ)Vq=r2kql.......(1)Due to 2q:V2q=r2k⋅2q⋅cos(120∘)⋅l=r28kql⋅(−21)=r2−4kql......(2)Using equations (1) and (2):Vnet=r2kql−r24kql=r2−3kqlVnet=−3(r2ql)×9×109Vnet=−27(r2ql)×109Vα=27
Q-9. If two charges q1andq2 are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?
Q-10. A stream of a positively charged particles having mq=2×1011,kgc and velocity vo=3×107i^m/s is deflected by an electric field 1.8 k V/m. The electric field exists in a region of 10 cm along x direction. Due to the electric field, the deflection of the charge particles in the y direction is __________mm.
Solution: 2 mm
a=mF=mqE=(2×1011)(1.8×103)=3.6×1014m/s2Time to cross plates:t=v0d=3×1070.10=3×1081sy=21at2=21(3.6×1014)(3×1081)2=9×1021.8=0.002m=2mm
Q-11. The electric potential at the centre of two concentric half rings of radii R1 and R2, having same linear charge density is
(1)ε02λ(2)2ε0λ(3)4ε0λ(4)ε0λ
Solution: 2ε0λ
Potential at the centre
V=4πε0R2(λπR1)+4πε0R1(λπR2)=2ε0λ
Q-12. In a cuboid of dimension 2L ✕ 2L ✕ 2L ,a charge q is placed at the centre of the surface 'S' having area of 4L2. The flux through the opposite surface to 'S' is given by
(1)12ε0q(2)3ε0q(3)4ε0q(4)6ε0q
Flux through whole cube=ε0qAs total faces are 6 so flux through ABCD isϕ=6ε0q
Q-13. A long cylindrical volume contains a uniformly distributed charge of density .The radius of cylindrical volume is R.A charge particle q revolves around the cylinder in a circular path. The Kinetic of the particle is
(1)ε0ρqR2(2)2ε0ρqR2(3)4ε0R2qρ(4)qρ4ε0kR2
Solution: ε0ρqR2
qenc=ρπr2lE⋅2πRl=ε0qenc=ε0ρπR2l⇒E=ε0ρRFor a particle of charge q revolving at radius R, the centripetal force is provided by electric field:qE=Rmv2q⋅ε0ρR=Rmv2⇒21mv2=2ε0qρR2K.E=4ε0qρR2
Q-14. In the figure a very large plane sheet of positive charge is shown P1andP2 are two points at distance l and 2l from the charge distribution.If σ is the surface charge density,then the magnitude of electric Fields E1andE2atP1andP2 respectively are :
(a)E1=ε0σ,E2=2ε0σ(b)E1=ε02σ,E2=ε0σ(c)E1=E2=2ε0σ(d)E1=E2=ε0σSolution:(c)E1=E2=2ε0σFrom a uniformly charged infinite sheet:E=2ε0σ.......(1)From the above equation it is clear that the electric field due to a uniformly charged surface is independent of distance from the surface.It only depends on surface charge density.Here in both the cases electric field will be the same(A)E1=E2=2ε0σ
Q-15. If a charge q is placed at the centre of a closed hemispherical non-conducting surface,the total flux passing through the flat surface would be
(1)ε0q(2)2ε0q(3)4ε0q(4)πε02qSolution:(2)2ε0qTotal flux through complete spherical surface isε0qSo the flux through curved surface is2ε0qThe flux through flat surface is zero
Table of Contents
1.0Key Concepts to Remember
1.1Definitions
1.2Important Formulas
1.3Short Tricks
2.0JEE Main Past Year Questions with Solutions on Electrostatics