Physics Electrostatic Potential

POTENTIAL DUE TO AN ELECTRIC DIPOLE

Electric fi eld at any point `A(r, theta)` due to dipole :

`V = V_+ + V_- = (kq)/r_+ + (k(-q))/(r_-) = (kq)/(r -lcostheta) + (k(-q))/(r+lcostheta)`

`= (k(2lq)) /(r^2 - l^2costheta^2)`

Solving and after using dipole approximation `l < < r`

`V = (kp)/r^2`

POTENTIAL DUE TO A SYSTEM OF CHARGES

Consider a system of charges `q_1, q_2,-, q_n` with position vectors `r_1, r_2,-, r_n` relative to some origin. The potential `V_1` at P due to the charge `q_1` is

`V_1 = 1/(4piepsilon_0) q_1/(r_(1P))`

where `r_(1P)` is the distance between `q_1` and P. Similarly, the potential `V_2` at P due to `q_2` and `V_3` due to `q_3` are given by

`V_2 = 1/(4piepsilon_0) q_2/(r_(2P))` `V_3 = 1/(4piepsilon_0) q_3/(r_(3P))`

where `r_(2P)` and `r_(3P)` are the distances of P from charges `q_2` and `q_3,` respectively; and so on for the potential due to other charges. By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges

`V = V_1 + V_2 + ... + V_n`

`V_2 = 1/(4piepsilon_0)(q_1/r_(1P)+ q_2/(r_(2P)) +......+ q_n/(r_(nP)))`

the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by

`V = 1/(4piepsilon_0)(q/r)` `(r>=R)`
The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell).

 
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