Dear Gajendra,

A charge q is moved from P₀ to P₁ in the vicinity of charge q (see Figure 25.1). The electrostatic potential at P₁ can be determined using eq. (25.4) and evaluating the integral along the path shown in Figure 25.1. Along the circular part of the path the electric field and the displacement are perpendicular, and the change in the electrostatic potential will be zero. Equation (25.4) can therefore be rewritten as

(25.5)

If the charge q is positive, the potential increases with a decreasing distance r. The electric field points away from a positive charge, and we conclude that the electric field points from regions with a high electrostatic potential towards regions with a low electrostatic potential.

Figure 25.1. Path followed by charge q between P₀ and P₁.

From the definition of the electrostatic potential in terms of the potential energy (eq.(25.3)) it is clear that the potential energy of a charge q under the influence of the electric field generated by charge q is given by

(25.6)

Example: Problem 25.21

 A total charge Q is distributed uniformly along a straight rod of length L. Find the potential at point P at a distance h from the midpoint of the rod (see Figure 25.2).

The potential at P due to a small segment of the rod, with length dx and charge dQ, located at the position indicated in Figure 25.3 is given by

(25.7)

The charge dQ of the segment is related to the total charge Q and length L

(25.8)

Combining equations (25.7) and (25.8) we obtain the following expression for dV:

(25.9)

Figure 25.2. Problem 25.21.

Figure 25.3. Solution of Problem 25.21.

The total potential at P can be obtained by summing over all small segments. This is equivalent to integrating eq.(25.9) between x = - L/2 and x = L/2.

(25.10)

Example: Problem 25.15

 An alpha particle with a kinetic energy of 1.7 x 10^-12 J is shot directly towards a platinum nucleus from a very large distance. What will be the distance of closest approach ? The electric charge of the alpha particle is 2e and that of the platinum nucleus is 78e. Treat the alpha particle and the nucleus as spherical charge distributions and disregard the motion of the nucleus.

The initial mechanical energy is equal to the kinetic energy of the alpha particle

(25.11)

Due to the electric repulsion between the alpha particle and the platinum nucleus, the alpha particle will slow down. At the distance of closest approach the velocity of the alpha particle is zero, and thus its kinetic energy is equal to zero. The total mechanical energy at this point is equal to the potential energy of the system

(25.12)

where q₁ is the charge of the alpha particle, q₂ is the charge of the platinum nucleus, and d is the distance of closest approach. Applying conservation of mechanical energy we obtain

(25.13)

The distance of closest approach can be obtained from eq.(25.13)

(25.14)

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u=q/4π εr

where ε is relative permitivity of air medium.