Electrostatics is the branch of physics that deals with the study of interactions between the stationary and slow-moving charges. Electric Force problems can be a bit tricky to solve. At first glance, these problems can seem to be scary and overwhelming to some students but no need to worry because this article is all about how you can easily solve electrostatics problems. 

• The major topics under this chapter are Coulomb’s Law, Electric Field, Dipole, etc. Check JEE Main Physics Syllabus
• The questions from this chapter will also be interrelated to other topics and can also be solved by using the concepts of work, energy, force, etc.

JEE Main Exam PatternJEE Main Chemistry Syllabus JEE Main Exam Dates

First of all, candidates need to read the problems or questions carefully so that they can identify all the known and unknown variables and this will be possible only if you have practiced and revised enough during your preparations. Moreover, candidates can expect one to two questions from this section in the examination. Go through the revision notes to earn your four marks from this section. 

Must Read: 

• JEE Main Study Notes for Work Energy and Power
• JEE Main Study Notes for Current Electricity
• JEE Main Study Notes on Laws of Motion

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Electric Charges and their Conservation

Electric changes are one of the fundamental physical properties of matter due to which it experiences a force when kept in the Electromagnetic field. Three types of changed particles are:

• Positive Charge (Proton)
• Negative Charge (Electron)
• Neutral Charge (Neutron)

Conservation of Charge

For an isolated system, the net charge of the system remains constant. Therefore charge cannot be created and cannot be destroyed. 

Properties of Electic Charge

• The electric charge is quantized. 

Q = ne

e = 1.602 X 10-19 Coulomb

• The like charges repel each other and unlike charges attract each other. 

Sample Question

Question: Let us consider that a hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. Calculate the potential at a distance of 2 cm from the center of the sphere?

1. zero
2. 10 V
3. 4 V
4. 10/3 V

Solution: (2)

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Coulomb’s Law

Coulomb’s Law states that the force acting between the two charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. 

Coulombs can be represented by 

Where is the constant of proportionality and is the permittivity of free space. 

Vector Form of the Coulomb’s Law is

Sample Question

Question: Let us consider that Two equally charged spheres of radii a and b are connected together. Calculate the ratio of electric field intensity on their surfaces?

1. a/b
2. a2/b2
3. b/a
4. b2/a2

Solution: (3)

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Superposition Principle

By using the superposition principle the force between the multiple charges can be found.  According to the superposition principle, the net force on any charge due to multiple charges is the vector sum of all the forces acting on that charge due to the other charges are taken at a time. The individual forces remain unaffected due to other charges presence. 

For n number of charges, the net resultant force due to the principle of superposition can be represented by 

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Continuous Charge Distribution

When a charge is continuously distributed over a surface of the conductor then it is known as Continuous Charge Distribution. The space and time of the charges in the continuous charge system is insignificant. 

Types of Continuous Charge Distribution

Line/ Linear Charge Distribution: When the charges are consistently spreaded over the length L of a conductor then it is called Line Charge Distribution. It is represented by λ. 

λ = dq/dl Cm–1

Area/ Surface Charge Distribution: When the charge is consistently spread over the surface area of a conductor then it is known as Surface Charge Distribution. It is represented by σ

σ = dq/ds Cm–2

Volume Charge Distribution: When the charge is consistently distributed over the volume of a conductor then it is known as Volume Charge Distribution. It is represented by ρ

ρ = dq/dV Cm–3

Sample Question

Question: Which of the following figure shows the correct equipotential surfaces of a system of two positive charges?

Solution: (c)

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Electric Field and Dipole

Electric Filed: Electric Field is defined as the force experienced per unit positive test charge at a point when another charge is kept in the vicinity.

Electric Dipole and Electric Dipole Moment: Electric Dipole is a system of two equal in magnitude opposite changes that are separated by a very small distance. 

Electric Dipole Moment is the product of charge and separation between them. It is a vector quantity. 

Electric Filed due to a Dipole

• On Axial Point: Electric field due to dipole on the axial point P is represented by
• On Equatorial Plane: Electric field due to the dipole on the equatorial plane at a point P is resented by
• Torque experienced by a dipole in a uniform electric field is represented by

Sample Question

Question: Let us consider that From a point charge, there is a fixed point A. At point A, there is an electric field of 500 V/m and a potential difference of 3000 V. Calculate the Distance between the point charge and A?

1.  6 m
2. 12 m
3. 16 m
4. 24 m

Solution: (1)

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Electric Flux

The number of electric lines of force crossing a surface normal to the area gives the electric flux ϕE

The electric flux through an elementary area ds is defined as the scalar product of area and field. 

d ϕE = Eds cos θ

OR

ϕE = ∫ E. ds

• Electric Flux will be maximum when an electric field is normal to the area d ϕ = Eds
• Electric Flux will be minimum when the field is parallel to area d ϕ = t0
• For a closed surface, inward flux is negative and outward flux is positive.

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Electric Potential

The electric potential at a point in a field is the amount of work done in bringing a unit positive charge from infinity to the point. It is equal to the Electric potential energy of unit positive charge at that point.

• It is a scalar
• S.I unit is volt

Electric Potential at a distance ‘d’ due to a point charge q in air or vacuum is represented by V.

• The gain in Kinetic Energy is: ½ mv2 = qV
• The gain in Velocity is v = √ 2qV/m

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Equipotential Surface

It is a surface on which all points are at the same potential.

1. The electric field is perpendicular to the equipotential surface
2. Work done in moving a charge on the equipotential surface is zero.

In the case of a Hollow Charged Sphere.

1. Intensity is zero at any point inside the sphere.
2. Intensity at any point on the surface is the same and it is maximum
3. Eclectic field intensity in vector form is represented by

In the Case of Solid Charged Sphere

The potential at any point inside the sphere and at any point on its surface is the same 

It is an equipotential surface. Outside the sphere, the potential varies inversely as the distance of the point from the centre.

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Electron volt

Electron Volt is the unit of energy in particle physics and is represented as eV.

• 1 eV = 1.602×10‑19 J.

Charged Particle in Electric Field

When a positive test charge is fired in the direction of an electric field the following occurs

• It accelerates,
• Its kinetic energy increases and hence
• Its potential energy decreases.

A charged particle of mass m carrying a charge q and falling through a potential V acquires a speed of

Sample Question

Question: The expression E=−dv/dr implies that the electric field is in that direction in which

1. An increase in potential is steepest.
2. A decrease in potential is steepest.
3. change is potential is minimum.
4. none of these

Solution: (2)

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Electric Lines of Force

The electric line of force is the path along which a unit positive charge accelerates in the electric field and the tangent at any point to the line of force gives the direction of the field at that point.

Properties of Electric Lines of Force

• Two lines of force can never intersect with each other.
• The number of lines of force passing normally through a unit area around a point is numerically equal to E and to the strength of the field at the point.
• On a charged conductor lines of force always leave or end normally.
• Electric lines of force can never be in closed loops.
• Lines of force have a tendency to contract longitudinally and exert a force of repulsion on one another laterally.
• There will be no lines of force if in a region of space there is no electric field. 
• There cannot be any line of force inside a conductor.

The number of lines of force passing normally through a unit area around a point is numerically equal to E.

In a uniform field, lines of force are parallel to one another.

Difference between magnetic lines of force and electric lines of force

• Electric lines of force can never form closed loops whereas magnetic lines are always in closed loops.
• Magnetic lines of force may exist inside a magnetic material but electric lines of force do not exist inside a conductor.

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Conductors, Insulators, and Semiconductors

• A body in which electric charge can easily flow through it is called conductor (e.g. metals).
• An insulator is a body in which an electric charge cannot flow. It is also called a dielectric. (e.g. glass, wool, rubber, plastic, etc.)
• Substances that are intermediate between conductors and insulators are called semiconductors. (e.g. silicon, germanium, etc)

Dielectric Strength: Dielectric Strength is the minimum field intensity that should be applied to break the insulating property of the insulator.

• Dielectric strength of air = 3 106 V/m
• Dielectric strength of Teflon = 60 × 106 Vm–1

The maximum charge that a sphere can hold totally depends on the size and dielectric strength of the medium in which the sphere is placed.

1. A maximum charge a sphere with a radius of ‘r’ can hold in air = 4ε0r2 dielectric strength of air.
2. If the electric field in the air exceeds its dielectric strength air molecules then it becomes ionized and are accelerated by fields and the air becomes conducting.

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Tips to Solve Questions on Electrostatics

1. Question: Charge +q and -q are placed at points A and B respectively which are at a distance of 2L apart, C is the midpoint between A and B. Calculate the work done in moving a charge +Q along the semicircle CRD is 

1. qQ/2πε0L
2. qQ/3πε0L
3. -qQ/6πε0L
4. qQ/4πε0L

Tip to Solve the Question: First of all you have to calculator the values of potentials Vc and Vd, then you can calculate the work of work.

Solution: (3)

2. Question: If there is an infinite straight chain of alternating charges q and -q. It is given that the distance between the two neighboring charges is equal to a. Calculate the interaction energy of any charge with all the other charges.

1. 2q2/4πε0a
2. 2q2loge2/4πε0a
3. −2q2loge2/4πε0a
4. Zero

Tip to Solve the Question: You can solve this question by using the formula

Solution: (3)

3. Question: The work done in placing a charge of 8×10−18coulomb on a condenser of capacity 100 microfarad is

Tip to Solve the Question: This question is direct and you can easily solve it if you remember the formula of Work Done that is W = Work done = ½ q2/C

Solution: (3)

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45 Days Study Plan for JEE Main Physics

Candidates can divide their remaining 45 days to the exam into weeks and make the best strategic plan to cover all the topics of JEE Main 2020. We have also prepared a study plan that you can apply or modify as per your convenience. 

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Tricks to Solve Question on Electrostatics

• Try to draw a diagram that represents all the charges in the problem.
• You have to identify the charges of interest and q. This can be only possible if you have done proper practice for this section. 
• Try to convert all units to SI.
• Identify and covert the charges in coulombs and distances in meters.
• Maintain consistency with the SI value of the Coulomb constant.
• Wherever it is necessary first of all apply Coulomb’s law and for each charge Q , find the electric force on the charge of interest, q. 
• Find the magnitude of the force using Coulomb’s law.
• Sum all the x-components to get the x-component of the resultant electric force.
• Sum all the y-components to get the y-component of the resultant electric force.