IIT JEE Physics Formula Revision 29. Electric Field and Potential

For revision points of the chapter

http://iit-jee-physics.blogspot.com/2008/03/concept-review-ch-29-electric-field-and.html


The charge on a proton is
e = 1.60218*10^-19 C



Coulomb's formula for the electric force between two charges.

(1) F = k*q1*q2/r²

q1,q2 charges
r = separation between charges
k = constant
In SI units k is measured to be 8.98755*10^9 N-m²/C²

The constant k is often written as 1/4πε₀.
The constant ε₀ is called the permittivity of the space and its value is

ε₀ = 8.85419*10^-12 C²/N-m²

The intensity of field is defined as Vector E = Vector F/q

(3) E = F/q

Electric field due to a point charge (Q)

(4) E = F/q = Q/4πε₀r²



(5) U(r2) - U(r1) = -W = [q1*q2/4πε₀]* [(1/r2) - (1/r1)]

Choosing potential energy at infinite separation as zero

(6) U(r) = U(r) - U(∞) = [q1*q2/4πε₀]* (1/r)
= q1*q2/4πε₀r

(7) VB - VA = (UB - UA)/q

We can define absolute electric potential at any point by choosing a reference point P and saying that the potential at this point is zero.

(8) VA = VA - VP = (UA - UP)/q

The potential due to a point charge Q (placed at A) at a point P with distance AP = r is

(9) VP = Q/4πε₀r
The electric potential at a point due to more than one charge in the system is obtained by finding potential due to individual charges and then adding them.

10. Relation between electric field and potential

dV = -E.dr

Integrating between limits r1 and r2

V2 – V1 = - ∫ E.dr

If we choose r1 as infinity (reference point)

V(r) = - ∫ E.dr (from infinity to r)

Hence we can calculate potential V if we know E and r.

If we know V we can find E through the relation

Ex = - ∂V/∂x
Ey = -∂V/∂y
Ez = -∂V/∂z

We can find the x,y and z components of E.

E can be written as Ex i +Ey j + Ez k

We can write dV as dV = -E dr cos θ where θ is the angle between the field E and the small displacement dr.

-dV/dr = E cos θ

15. Electric Dipole

Electric dipole moment
It is defined as a vector p = q*distance vector d
where distance vector is the vector joining the negative charge to the positive charge.
The line along the direction of the dipole moment is called the axis of the dipole.

16. Electric potential due to a dipole at a point P

Point is at distance r from the centre of the diploe (d/2) and theline joining the point P to the centre of the dipole make an angle θ with the direction of dipole movement (from –q to +q)

Potential at P due to charge –q = - [1/4πε₀][q/(r + (dcos θ)/2)]
Potential at P due to charge q = [1/4πε₀][q/(r - (dcos θ)/2)]
Net potential due to q and –q = [1/4πε₀](qd cos θ)/r²

17. Electric field due to a dipole

E_r = [1/4πε₀](2p cos θ)/r³

E_θ = [1/4πε₀](p sin θ)/r³

Resultant electric field at P = E= √ (E_r²+ E_θ²)

= [1/4πε₀](p/r³)√(3 cos²θ + 1)


19. Torque on an electric dipole placed in an electric field.

If the dipole axis makes an angle θ with the electric field magnitude of the torque = | Γ| = pE sin θ

In vector notation Γ = p × E


20. Potential energy of a dipole placed in a uniform electric field

dipole axis makes an angle θ with the electric field magnitude of the torque

Change in potential energy = U(θ) – U(90°) = -pE cos θ = -p.E

If we choose the potential energy of the dipole to be zero when θ = 90° , above equation becomes
U(θ) = -pE cos θ = -p.E