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JEE Physics
Electrostatics

Electrostatics

Electrostatics pertains to the study of stationary electric charges and their associated phenomena and properties in physics.. It primarily focuses on the behavior of these charges and their interactions at rest without considering the effects of their motion (which falls under electrodynamics). This may include the study of electric Charge, coulomb's law, electric field, electric Potential, gauss law, electrostatics potential energy, capacitors.

1.0Electric charge

Charge is an inherent characteristic of matter that generates and interacts with electric and magnetic phenomena..

Types of Charges

(1). Positive charge- Due to deficiency of electrons.

(2). Negative charge-Due to excess of electrons

Units of Charge

  • SI - Coulomb(C)
  • CGS-Stat coulomb (stat C) 
  • Dimensional Formula -[AT]

2.0Basic Properties of Electric charge

(1). Additivity of electric Charge.

(2). Conservation of electric Charge

(3). Quantization of Electric Charge

(4). Charge can be transferred

(5). Charge is invariant.

(6). Charge is a scalar quantity

(7). Charge is always associated with mass

Methods of charging

(1). Charging by Friction

(2). Charging by Induction

(3). Charging by Conduction

3.0Coulomb’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.

Coulomb's law


F=Kr2q1​q2​​

k=4πϵ0​1​=9×109Nm2C−2 = Electrostatic constant or Coulomb’s Constant

Coulomb's Law in Vector Form

Coulomb's law in vector form


F21​​=4πϵ0​1​q1​q2​∣r2​−r1​3(r2​​−r1​)​ or 

​F12​=4πϵ0​1​q1​q2​∣r1​​−r2​​∣3r1​​−r2​​​

4.0Superposition Principle

When several charges interact, the total force on a given charge is the vector sum of all the individual forces exerted on it by all other charges.

Superposition Principle


F=F01​​+F02​​+F03​​+…..F0n​​

F=r12​kq0​q1​​r1​​+r22​kq0​q2​​r2​​+……+ri2​kq0​qi​​ri​​+….rn2​kq0​qn​​rn​​

Relative Permittivity (ϵr​) or Dielectric Constant (K)

Dielectric Constant (K) =  Permittivity of Vacuum(ϵ0​) Permittivity of a Medium (ϵr​)​ 

K=ϵ0​ϵr​​

ϵ=ϵ0​ϵr​

The value of ϵr​ or K≥1.

5.0Electric Field Intensity

  • The space around a charge or charge distribution, where another charge experiences an electric force, is called an electric field.
  • Mathematically, E=q0​F​ and SI Unit : N/C or V/m
  • E=limq0​→0​q0​F​
  • Dimensional Formula: [M1L1T−3A−1]
  • It is a vector quantity.

6.0Electric Field Lines 

  • Electric field lines are conceptual lines, either straight or curved, used to represent the direction and strength of the electric field surrounding an electrically charged object. a region such that the tangent at any point on the field lines gives the direction of the electric field lines at that point in the area.
  • Two electric field lines can never intersect each other.

Electric field lines

  • Electrostatic field lines can never form any closed loop

Electric field lines due to positive or negative charges

Electric field lines due to positive or negative charges

7.0Electric line of force due to an electric dipole

Electric line of force due to an electric dipole

8.0Continuous Charge Distribution

A collection of closely positioned electric charges constitutes a continuous charge distribution.

(1). Linear Charge Density (λ): λ=lQ​

 SI Unit mC​

(2). Surface Charge Density (σ):  σ=SQ​

 SI Unit m2C​

(3). Volume Charge Density  ( ρ ) 

ρ=VQ​

 SI Unit m3C​

9.0Electric Dipole

It is a pair  of two same  and opposite charges placed at small separation.

Electric Dipole Moment


  • The dipole moment equals the product of the magnitude of either Charge and the separation between the charges: p​=q(2l)
  • S.I. unit- Cm

Electric Field Due to a Dipole

(1) At Axial / End on position

Electric Field Due to a Dipole at axial position


EAxial ​=4πϵ0​r312p​

(2) At Equator/Broadside on position

Electric Field Due to a Dipole at equator

EEqui ​=4πϵ0​r31p​

(3) At general position

Electric dipole moment - general case


E=4πϵ0​r3p​3cos2ϑ+1​

Tanα=21​Tanθ

Electric Field Intensity due to a charged wire

Electric Field Intensity due to a charged wire


Special Cases:

  1. For Infinite Wire, ( both ends goes to infinite)

EX​=E⊥​=r2Kλ​

EY​=E∥​=0

  1. For Semi-infinite wire

For Semi-infinite wire

EX​=E⊥​=rKλ​

EY​=E∥​=rKλ​

EResultant ​=2​rKλ​

  1. Electric field due to finite wire at symmetric point:

Ex​=E⊥​=r2Kλ​sinθ

EY​=E∥​=0

  1. Electric Field due to a uniformly charged Arc

Electric Field due to a uniformly charged Arc


E=R2Kλ​Sin(2θ​)

  1. Electric field at centre of uniformly charged Ring

E=R2Kλ​Sin(2θ​),θ→ angle of arc 

θ=360∘

E=R2Kλ​Sin(2360​)

E = 0

  1. Electric Field due to a Uniformly charged Ring at its Axis

Electric Field due to a Uniformly charged Ring at its Axis


E=(R2+x2)3/2KQx​

Special cases:

  • Electric field on the axis for small values of x: E=R3KQ​⋅X
  • Electric field at the centre of the ring is zero because x = 0
  • Electric field at the axis for larger values of x: E≃X2kQ​
  • Maximum value of electric field

dxdE​=0

x=∓2​R​

Emax ​=33​R22kQ​

Variation of electric field with distance


10.0Electrostatic Potential Energy(EPE)

  • Electrostatic Potential Energy, is the amount of work done in bringing the charge from infinity to the present position without changing its kinetic energy.

Electrostatic Potential Energy (EPE) of two point charge system 

Electrostatic Potential Energy (EPE) of two point charge system

ΔU=(W∞→r​)ext ​=−∫∞r​Fext ​⋅dr

⇒Ur​−U∞​=−∫∞r​Fext ​⋅dr=rkQq​

(∴U∞​=0)

⇒U−0=rkQq​

⇒U=rkQq​

Electrostatic Potential energy of a three system of charges

Electrostatic Potential energy of a three system of charges


US​=U12​+U23​+U13​

US​=r12​kq1​q2​​+r23​kq2​q3​​+r13​kq1​q3​​

Note: For n point charges system 

No. of pairs =2n(n−1)​

11.0Electric 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(w∞→p​)ext​​(ΔK=0)

  • It is scalar quantity
  • Potential can be positive, negative and even zero
  • SI unit is joule/coulomb

Electrostatic potential due to a point charge

  • V=rkQ​

Expression for potential due to multiple Charges

Expression for potential due to multiple Charges

V=V1​+V2​+V3​+……..Vn​

V1​=4πϵ0​1​r1P​q1​​,V2​=4πϵ0​1​r2P​q2​​,V3​=4πϵ0​1​r3P​q3​​

12.0Equipotential Surface(E.P.S)

Locus of all such points which are at the same potential is called equipotential surface.

Point charge

Equipotential Surface of a point charge


Plane Charge Sheet

EPS for plane sheet

Unlike Charges(Electric Dipole)

EPS for Unlike Charges(Electric Dipole)

Like Charges(Similar Charges)

EPS for Like Charges(Similar Charges)

13.0Solved Questions on Electrostatics

Q-1 Two spheres having equal charges are kept at long separation. If the gravitational force between two spheres is equal to electrostatic force between them. Find the ratio of specific charge mq​.

Solution:

 r2kq2​=r2Gm2​

 mq​=kG​​=9×1096.67×10−11​​=0.86×10−10

Q-2 Two thin spherical shells made of metal are at a larger distance apart . One of radius 10 cm carries a charge of +0.5 mC and the other, of radius 20 cm , carries a charge of + 0.7 mC. The charge on each, when they are connected by a suitable conducting wire is respectively.

Solution:

Q1​=(R1​+R2​R1​​)Q=(30 cm10 cm​)1.2mC⇒Q1​=0.4mC               

 Q1​=(R1​+R2​R2​​)Q=(30 cm20 cm​)1.2mC⇒Q2​=0.8mC

Q-3. The electric field in a certain region is given by E=(x3K​)i^. The dimensions of K are.

Solution:

Dimension of electric field = [M1L1T--3A-1]

E=(x3K​)i^

E=x3K​

K=Ex3

( ∴ x representing  the distance so the dimensional formula is [L]

K=[M1L1T--3A-1] [L]3

K=[M1L4T-3A-1]

Q-4. If the electric flux entering and leaving a closed surface is ϕ1​ and ϕ2​ respectively then electric charge inside the surface will be.

Solution:

According to Gauss Theorem, ϕ=ϵ0​q​

Net Flux (ϕ)=ϕ2​−ϕ1​                      

ϕ2​−ϕ1​=ϵ0​q​           

 q=(ϕ2​−ϕ1​)ϵ0​                 

Q-5. The electric potential and electric field at a point due to a point charge are 600 V and 200 N/C respectively. Calculate the magnitude of point Charge.

Solution:

Electric Potential (V)=rkq​=600 ……..(1)

Electric Field (E)=r2kq​=200 ……..(2)

Squaring both the sides in eq.(1)

r2k2q2​=360000 ………(3) 

From eq (2) and (3)

 kq= 1800

q=k1800​=9×1091800​=200×10−9c

Table Of Contents:


  • 1.0Electric charge
  • 1.1Types of Charges
  • 1.2Units of Charge
  • 2.0Basic Properties of Electric charge
  • 2.1Methods of charging
  • 3.0Coulomb’s Law
  • 3.1Coulomb's Law in Vector Form
  • 4.0Superposition Principle
  • 5.0Electric Field Intensity
  • 6.0Electric Field Lines 
  • 6.1Electric field lines due to positive or negative charges
  • 7.0Electric line of force due to an electric dipole
  • 8.0Continuous Charge Distribution
  • 9.0Electric Dipole
  • 9.1Electric Field Due to a Dipole
  • 9.2Electric Field Intensity due to a charged wire
  • 10.0Electrostatic Potential Energy(EPE)
  • 10.1Electrostatic Potential Energy (EPE) of two point charge system 
  • 10.2Electrostatic Potential energy of a three system of charges
  • 11.0Electric potential
  • 12.0Equipotential Surface(E.P.S)
  • 12.1Point charge
  • 12.2Unlike Charges(Electric Dipole)
  • 12.3Like Charges(Similar Charges)
  • 13.0Solved Questions on Electrostatics

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