What is the importance of practising Electrostatics PYQs?

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.

Which subtopics from Electrostatics are most commonly appeared in the IIT JEE 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.

How many questions come from Electrostatics in JEE Mains?

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 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

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.

$F = K \frac{q _{1} q _{2}}{r ^{2}} k = \frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} Nm^{2} C^{- 2} = Electrostatic constant or Coulomb’s Constant$

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, $E = \frac{F}{q _{0}}, SI unit: N/C or V/m$

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.

$ϕ = \oint E \cdot d S = \frac{q _{enclosed}}{ε _{0}}$

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.

$V_{p} = \frac{( w _{on} - w _{off} )}{q}$

5. Electrical Capacitance(C)

It shows the capacity of a conductor to store electric charge.

$C = \frac{Q}{V}$

Important Formulas

Coulomb’s Law	$F = K \frac{q _{1} q _{2}}{r ^{2}} k = \frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} Nm^{2} C^{- 2} = Coulomb Constant$
Coulomb's Law in Vector Form	$F_{21} = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2} ( r _{2} - r _{1} )}{r _{2} - r _{1} ^{3}} O r F_{12} = - \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2} ( r _{2} - r _{1} )}{r _{2} - r _{1} ^{3}}$
Superposition Principle	$F = F_{01} + F_{02} + F_{03} + \dots + F_{0 n} F = \frac{k q _{0} q _{1}}{r _{1}^{2}} \overset{r}{^}_{1} + \frac{k q _{0} q _{2}}{r _{2}^{2}} \overset{r}{^}_{2} + \frac{k q _{0} q _{3}}{r _{3}^{2}} \overset{r}{^}_{3} + \dots + \frac{k q _{0} q _{n}}{r _{n}^{2}} \overset{r}{^}_{n}$

Short Tricks

Case-1:

The equilibrium position will be near to a smaller charge.

$x = \frac{r}{\frac{Q _{1}}{Q _{2}} + 1} Q_{1} Q_{2} > 0 and ∣ Q_{1} ∣ < ∣ Q_{2} ∣$

Case-2:

The equilibrium position will be near to a smaller charge

$x = \frac{r}{\frac{Q _{1}}{Q _{2}} - 1} Q_{1} Q_{2} < 0 and ∣ Q_{1} ∣ < ∣ Q_{2} ∣$

Electric Field Intensity	$E = lim_{q_{0} \to 0} \frac{F}{q _{0}}$
Linear Charge Density()	$λ = \frac{Q}{l}$
Surface Charge Density ()	$σ = \frac{Q}{S}$
Volume Charge Density()	$ρ = \frac{Q}{V}$
At Axial / End on position	$E_{Axial} = \frac{1}{4 π ε _{0}} \cdot \frac{2 p}{r ^{3}}$
At Equator/Broadside on position	$E_{Equi} = \frac{1}{4 π ε _{0}} \cdot \frac{p}{r ^{3}}$
At general position	$E = - \frac{p}{4 π ε _{0} r ^{3}} (3 cos^{2} θ + 1) tan α = \frac{1}{2} tan θ$
For Infinite Wire,( both ends goes to infinite)	$E_{x} = E_{⊥} = \frac{2 kλ}{r}$
For Semi-infinite wire	$E_{x} = E_{1} = \frac{kλ}{r}, E_{y} = E_{2} = \frac{kλ}{r} E_{Resultant} = 2 \cdot \frac{kλ}{r}$
Electric Field due to a uniformly charged Arc	$E = \frac{2 kλ}{R} sin (\frac{θ}{2})$
Electric field at centre of uniformly charged Ring	E=0
Electric Field due to a Uniformly charged Ring at its Axis	$E = \frac{k Q x}{( x ^{2} + r ^{2} ) ^{3/2}}$
Dipole placed in an electric field,	$τ = pE sin θ$
Work done in rotating a dipole in a uniform electric field	$W = pE (cos θ_{1} - cos θ_{2})$

Electric flux through a large surface	$Φ = \int d ϕ = \int E \cdot d A$
Gauss Law-	$Φ = \oint E \cdot d S = \frac{q _{enclosed}}{ε _{0}}$
Electric field Intensity due to infinitely long wire	$E = \frac{λ}{2 π ε _{0} r} = \frac{2 kλ}{r}$
Electric field due to uniformly charged long cylindrical pipe/cylindrical shell	$(a) E = \frac{σ R}{ε _{0} r} (r > R) (b) E = \frac{σ}{ε _{0}} (r = R) (c) E_{inside} = 0 (r < R)$
Electric Field due to Uniformly Charged Infinite Sheet	$(a) E = \frac{σ}{2 ε _{0}} (Non Conducting Sheet) (b) E = \frac{σ}{ε _{0}} (Conducting Sheet)$
Electric Field due to the charged conducting sphere or charged thin shell	$E = \frac{q}{4 π ε _{0} r ^{2}} = \frac{k q}{r ^{2}} (r > R) E_{s} = \frac{k q}{R ^{2}} = \frac{q}{4 π ε _{0} R ^{2}} = \frac{σ}{ε _{0}} (since σ = \frac{q}{4 π R ^{2}}) (r = R) E_{inside} = 0 (r < R)$
.Electric Field due to uniformly charged non conducting sphere(solid sphere)	$E = \frac{k q}{r ^{2}} (r > R) E_{s} = \frac{k q}{R ^{2}} = \frac{q}{4 π ε _{0} R ^{2}} = \frac{ρR}{3 ε _{0}} (r = R) E_{inside} = \frac{1}{4 π ε _{0}} \cdot \frac{q r}{R ^{3}} = \frac{ρ r}{3 ε _{0}} (r < R)$

Electrostatic Potential Energy(EPE)	$U = \frac{k Qq}{r}$
EPE For Three system of charges	$U = \frac{k q _{1} q _{2}}{r _{12}} + \frac{k q _{2} q _{3}}{r _{23}} + \frac{k q _{1} q _{3}}{r _{13}}$
Electric potential	$V_{p} = \frac{( w _{on} - w _{off} )}{q}, (Δ K = 0), V = \frac{k Q}{r}$
Electrical Capacitance(C)	$C = \frac{Q}{V}$
Capacitance of Parallel Plate Capacitor	$C = \frac{ε _{0} A}{d}$
Capacitance Of Spherical Capacitor	$C = 4 π ε_{0} r C = 4 π ε_{0} (\frac{ab}{b - a})$
Capacitors in series	$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}} + \dots + \frac{1}{C _{n}}$
Capacitors in parallel	$C = C_{1} + C_{2} + C_{3} + \dots + C_{n}$
Energy stored in a capacitor	$U = \frac{1}{2} C V^{2} U = \frac{1}{2} Q V U = \frac{1}{2} \frac{Q ^{2}}{C}$
Energy Density in a parallel plate capacitor	$U_{E} = \frac{1}{2} ε_{0} E^{2}$
Common Potential	$V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$
Loss of Heat Energy on sharing charges	$Heat = \frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )}$
If unlike plates are connected,heat loss	$Heat = \frac{C _{1} C _{2} ( V _{1} + V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )}$
Capacitance of parallel plate when dielectric is partially filled	$C = \frac{ε _{0} A}{( d - t ) + \frac{t}{k}}$
Force between plates of parallel plate capacitor	$F = \frac{Q ^{2}}{2 A ε _{0}} = \frac{2 d A - C V ^{2}}{2 d}$
Pressure on each plate of a capacitor	$P = \frac{σ ^{2}}{2 ε _{0}}$

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:

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle with each other.

Case-1 When suspended in water

$T cos θ_{1} = m g T sin θ_{1} = F Dividing both equations, we get, tan θ_{1} = \frac{F}{m g} ........... (1)$

Case-2 When suspense in water,value of electrostatic force may changes,

$F^{'} = \frac{F}{K} F^{'} = T sin θ_{1} ....... (2) m g - U = T cos θ_{1} (where U = Upthrust) σV g - ρ V g = T cos θ_{1} (σ - ρ) V g = T cos θ_{1} \dot{˙} ........ (3) Dividing (2) by (3): tan θ_{1} = \frac{F ^{'}}{( σ - ρ ) V g} Dividing (1) by (4): 1 = \frac{F}{m g} \times \frac{( σ - ρ ) V g}{F} = \frac{F}{σV g} \times \frac{( σ - ρ ) V g}{F} \Rightarrow 1 = K \times \frac{( σ - ρ )}{σ} \times \frac{1.5 - 1}{1.5} = K \times \frac{0.5}{1.5} = K \times \frac{1}{3} \Rightarrow K = 3 K = 3$

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 :

$(1) \frac{2 Q}{ε _{0}} (2) \frac{5 Q}{ε _{0}} (3) \frac{7 Q}{ε _{0}} (4) \frac{3 Q}{ε _{0}}$

Solution: (2) $\frac{5 Q}{ε _{0}}$

Two charges of 5Q and -2Q are situated at the points

5Q charge is inside the spherical region

$ϕ_{net} = \frac{q _{enclosed}}{ε _{0}} = \frac{5 Q}{ε _{0}}$

Q-3. An electric field is given by $(6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N/C$ . The electric flux through a surface area $30 \hat{i} m^{2}$ lying in YZ- plane (in SI unit) is

(1) 90

(2)150

(3)180

(4)60

Solution:

$E = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N/C A = 30 \hat{i} m^{2} ϕ = E \cdot A = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i}) = 6 \times 30 = 180$

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 $g c m^{- 3}$ , the angle remains the same. If density of material of the sphere is 1.6 $g c m^{- 3}$ , the dielectric constant of the liquid is_____

1) 1

2) 4

3) 3

4) 2

Solution: 42

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 density0.8 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_____

$T cos θ = m g T sin θ = F_{e} tan θ = \frac{F _{e}}{m g} tan θ = \frac{F _{e}}{ρ _{b} V g} ........ (1) tan θ = \frac{F _{e}}{( ρ _{b} - ρ _{l} ) V g} ...... (2) F ro m e q u a t i o n s (1) an d (2) : ρ_{b} V g = (ρ_{b} - ρ_{l}) kV g 1.4 = 0.7 k \Rightarrow k = 2$

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

$(1) T^{2} = \frac{4 π ^{2} m r ^{3}}{2 kλ q} (2) T = 2 π r \frac{m}{2 kλ q} (3) T = \frac{1}{2 π} \frac{m}{2 kλ q} (4) T = \frac{1}{2 π} \frac{2 kλ q}{m}$

Solution: $T = 2 π r \frac{m}{2 kλ q}$

$\frac{2 kλ q}{r} = m ω^{2} r ω^{2} = \frac{2 kλ q}{m r ^{2}} (\frac{2 π}{T})^{2} = \frac{2 kλ q}{m r ^{2}} T = 2 π r \frac{m}{2 kλ q}$

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

$(1) \frac{r}{1 + 3} (2) \frac{r}{3 ( 1 + 3 )} (3) \frac{r}{( 1 + 3 )}$

Solution: $\frac{r}{( 1 + 3 )}$

Two charges q and 3q are separated by a distance 'r' in the air.

$(E_{net})_{p} = 0 \frac{k q}{x ^{2}} = \frac{k \cdot 3 q}{( r - x ) ^{2}} \frac{1}{x ^{2}} = \frac{3}{( r - x ) ^{2}} \Rightarrow (r - x)^{2} = 3 x^{2} r - x = 3 x r = x (1 + 3) x = \frac{r}{1 + 3}$

Q-7. Force between two point charges $q_{1} an d q_{2}$ 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

$(1) F = \frac{25}{3} (2) 5 F (3) \frac{F}{5} (4) 25 F$

Solution:(2) 5F

In air, $F = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{r ^{2}}$

In medium, $F = \frac{1}{4 π ( k ε _{0} )} \cdot \frac{q _{1} q _{2}}{r ^{2}}$

$= \frac{q _{1} q _{2}}{4 π ε _{0} r ^{2}} \cdot \frac{1}{k} = \frac{25}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{r ^{2}} \Rightarrow 5 F$

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 $- α (\frac{ql}{r ^{2}}) \times 1 0^{9} V$ where the value of $α$ is

Solution:

$Due to q: V_{q} = \frac{k q c o s θ}{r} = \frac{k q ( 2 θ )}{r} V_{q} = \frac{k ql}{r ^{2}} ....... (1) Due to 2q: V_{2 q} = \frac{k \cdot 2 q \cdot c o s ( 12 0 ^{\circ} ) \cdot l}{r ^{2}} = \frac{8 k ql}{r ^{2}} \cdot (- \frac{1}{2}) = \frac{- 4 k ql}{r ^{2}} ...... (2) Using equations (1) and (2): V_{net} = \frac{k ql}{r ^{2}} - \frac{4 k ql}{r ^{2}} = \frac{- 3 k ql}{r ^{2}} V_{net} = - 3 (\frac{ql}{r ^{2}}) \times 9 \times 1 0^{9} V_{net} = - 27 (\frac{ql}{r ^{2}}) \times 1 0^{9} V α = 27$

Q-9. If two charges $q_{1} an d q_{2}$ 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?

$(1) \frac{a}{k} (2) k a (3) 1.5 a / k (4) 2 a / k$

$Solution: (1) \frac{a}{k} F = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{k d ^{2}} (in medium) F_{air} = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{d ^{2}} F = F_{air} = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{k d ^{2}} = \frac{1}{4 π ε _{0}} \cdot \frac{q _{1} q _{2}}{d ^{2}} d^{'} = \frac{d}{k}$

Q-10. A stream of a positively charged particles having $\frac{q}{m} = 2 \times 1 0^{11}, \frac{c}{k g}$ and velocity $v_{o} = 3 \times 1 0^{7} \hat{i} m/s$ 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 stream of a positively charged particles having and velocity is deflected by an electric field

$a = \frac{F}{m} = \frac{qE}{m} = (2 \times 1 0^{11}) (1.8 \times 1 0^{3}) = 3.6 \times 1 0^{14} m/s^{2} Time to cross plates: t = \frac{d}{v _{0}} = \frac{0.10}{3 \times 1 0 ^{7}} = \frac{1}{3 \times 1 0 ^{8}} s y = \frac{1}{2} a t^{2} = \frac{1}{2} (3.6 \times 1 0^{14}) (\frac{1}{3 \times 1 0 ^{8}})^{2} = \frac{1.8}{9 \times 1 0 ^{2}} = 0.002 m = 2 mm$

Q-11. The electric potential at the centre of two concentric half rings of radii R1 and R2, having same linear charge density is

$(1) \frac{2 λ}{ε _{0}} (2) \frac{λ}{2 ε _{0}} (3) \frac{λ}{4 ε _{0}} (4) \frac{λ}{ε _{0}}$

Solution: $\frac{λ}{2 ε _{0}}$

Potential at the centre

$V = \frac{( λπ R _{1} )}{4 π ε _{0} R _{2}} + \frac{( λπ R _{2} )}{4 π ε _{0} R _{1}} = \frac{λ}{2 ε _{0}}$

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 $4 L^{2}$ . The flux through the opposite surface to 'S' is given by

$(1) \frac{q}{12 ε _{0}} (2) \frac{q}{3 ε _{0}} (3) \frac{q}{4 ε _{0}} (4) \frac{q}{6 ε _{0}}$

$Flux through whole cube = \frac{q}{ε _{0}} As total faces are 6 so flux through ABCD is ϕ = \frac{q}{6 ε _{0}}$

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

$(1) \frac{ρq R ^{2}}{ε _{0}} (2) \frac{ρq R ^{2}}{2 ε _{0}} (3) \frac{qρ}{4 ε _{0} R ^{2}} (4) \frac{4 ε _{0} k R ^{2}}{qρ}$

Solution: $\frac{ρq R ^{2}}{ε _{0}}$

$q_{enc} = ρ π r^{2} l E \cdot 2 π Rl = \frac{q _{enc}}{ε _{0}} = \frac{ρ π R ^{2} l}{ε _{0}} \Rightarrow E = \frac{ρR}{ε _{0}} For a particle of charge q revolving at radius R, the centripetal force is provided by electric field: qE = \frac{m v ^{2}}{R} q \cdot \frac{ρR}{ε _{0}} = \frac{m v ^{2}}{R} \Rightarrow \frac{1}{2} m v^{2} = \frac{qρ R ^{2}}{2 ε _{0}} K . E = \frac{qρ R ^{2}}{4 ε _{0}}$

Q-14. In the figure a very large plane sheet of positive charge is shown $P_{1} an d P_{2}$ 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 $E_{1} an d E_{2} a t P_{1} an d P_{2}$ respectively are :

$(a) E_{1} = \frac{σ}{ε _{0}}, E_{2} = \frac{σ}{2 ε _{0}} (b) E_{1} = \frac{2 σ}{ε _{0}}, E_{2} = \frac{σ}{ε _{0}} (c) E_{1} = E_{2} = \frac{σ}{2 ε _{0}} (d) E_{1} = E_{2} = \frac{σ}{ε _{0}} Solution: (c) E_{1} = E_{2} = \frac{σ}{2 ε _{0}} From a uniformly charged infinite sheet: E = \frac{σ}{2 ε _{0}} ....... (1) From the above equation it is clear that the electric field due to a uniformly charged surface is independent of distance from the surface.It only depends on surface charge density. Here in both the cases electric field will be the same (A) E_{1} = E_{2} = \frac{σ}{2 ε _{0}}$

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

. 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

$(1) \frac{q}{ε _{0}} (2) \frac{q}{2 ε _{0}} (3) \frac{q}{4 ε _{0}} (4) \frac{2 q}{π ε _{0}} Solution: (2) \frac{q}{2 ε _{0}} Total flux through complete spherical surface is \frac{q}{ε _{0}} So the flux through curved surface is \frac{q}{2 ε _{0}} The flux through flat surface is zero$

Electrostatics Previous Year Questions with Solutions

1.0Key Concepts to Remember

Definitions

Important Formulas

Short Tricks

2.0JEE Main Past Year Questions with Solutions on Electrostatics

Table of Contents

Frequently Asked Questions

What is the importance of practising Electrostatics PYQs?

Which subtopics from Electrostatics are most commonly appeared in the IIT JEE exam?

How many questions come from Electrostatics in JEE Mains?

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