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.
F=Kr2q1q2
k=4πϵ01=9×109Nm2C−2 = Electrostatic constant or Coulomb’s Constant
Coulomb's Law in Vector Form
F21=4πϵ01q1q2∣r2−r13(r2−r1) or
F12=4πϵ01q1q2∣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.
Relative Permittivity (ϵr) or Dielectric Constant (K)
Dielectric Constant (K) = Permittivity of Vacuum(ϵ0) Permittivity of a Medium (ϵr)
K=ϵ0ϵr
ϵ=ϵ0ϵr
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=q0F and SI Unit : N/C or V/m
E=limq0→0q0F
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.
Electrostatic field lines can never form any closed loop
Electric field lines due to positive or negative charges
7.0Electric 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.
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
EAxial =4πϵ0r312p
(2) At Equator/Broadside on position
EEqui =4πϵ0r31p
(3) At general position
E=4πϵ0r3p3cos2ϑ+1
Tanα=21Tanθ
Electric Field Intensity due to a charged wire
Special Cases:
For Infinite Wire, ( both ends goes to infinite)
EX=E⊥=r2Kλ
EY=E∥=0
For Semi-infinite wire
EX=E⊥=rKλ
EY=E∥=rKλ
EResultant =2rKλ
Electric field due to finite wire at symmetric point:
Ex=E⊥=r2Kλsinθ
EY=E∥=0
Electric Field due to a uniformly charged Arc
E=R2KλSin(2θ)
Electric field at centre of uniformly charged Ring
E=R2KλSin(2θ),θ→ angle of arc
θ=360∘
E=R2KλSin(2360)
E = 0
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=∓2R
Emax =33R22kQ
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
ΔU=(W∞→r)ext =−∫∞rFext ⋅dr
⇒Ur−U∞=−∫∞rFext ⋅dr=rkQq
(∴U∞=0)
⇒U−0=rkQq
⇒U=rkQq
Electrostatic Potential energy of a three system of charges
US=U12+U23+U13
US=r12kq1q2+r23kq2q3+r13kq1q3
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.
Locus of all such points which are at the same potential is called equipotential surface.
Point charge
Plane Charge Sheet
Unlike Charges(Electric Dipole)
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+R2R1)Q=(30cm10cm)1.2mC⇒Q1=0.4mC
Q1=(R1+R2R2)Q=(30cm20cm)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=ϵ0q
Net Flux (ϕ)=ϕ2−ϕ1
ϕ2−ϕ1=ϵ0q
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