Potential Difference - Related to Source Charge

Description:

Let there is be a charge ‘Q’ creating an electric field all around it. Two points A and B are located on the electric field, such that points A and B are at a distance r_A and r_B respectively from the charge Q. Where r_AB. As shown in fig. So, Electric field at A will be stronger than electric field at B.

Since force at every point between A and B are different due to different electric field, so the work done on the charge will be calculated using calculus.

If P is a general point at distance ‘x’ from Q, Then electric field at P due to Q will be −

E_P = 1/4πε₀.Q/x²

If a charge is placed at point P and moved to a very small distance ‘dx’ at point P’ towards the direction A, some amount of work will have to be done. So the displacement of charge will be in the direction opposite to the direction of electric field. So, the angle between force on charge and the displacement of charge is 180.

Due to work done on charge, charge will experience some amount of force. Since the distance between the points A and B tends to zero, the force experienced by the charge will be same at both the points.

At point ‘P’ Force on charge q = q E

= 1/4πε₀.Qq/x²

Similarly, At point  P’ Force on charge q = q E

With changing force, it is not possible to get a unit value of Work at every point . Since the force is same at P and P’ so, Work done on charge ‘q’ = Force ∗ displacement.

Workdone for displacement dx = W = 1/4πε₀.Qq/x²dx.cos180

dW is the small work done due to very small displacement dx

So, dW = -1/4πε₀.Qq/x²dx

Work done to move the charge ‘q’ from distance r_B to r_A

Integrating above equation within the limit(r_B to r_A)

W_BA = r_A∫r_B -Qq/4πε₀.dx/x²

W_BA = -Qq/4πε₀ r_A∫r_B dx/x²

WBA = -Qq/4πε0[X-1/-1]rArB

⇒ W_BA = +Qq/4πε₀[1/r_A-1/r_B]

Work done per unit charge

W_AB/q = Q/4πε₀[1/r_A-1/r_B]

We know that Work done per unit charge is potential difference V_A-V_B.

So potential difference between two points (A and B)

V_A-V_B=Q/4πε₀[1/r_A-1/r_B]