Question

3. A charge q is uniformly distributed along an insulating straight wire of length 2 l as shown in

Fig. 24.34. Find an expression for the electric potential at a point located a distance d from the
distribution along its perpendicular bisector.
​

Answer 1

Given: A charge q is uniformly distributed along an insulating straight wire of length 2 l.

To find: An expression for the electric potential at a point located a distance d from the  distribution along its perpendicular bisector.

Solution:

• Consider a small element dx of the rod at distance x from mid point.

• Then Potential due to the small element dx will be:

                 dPE = k(qdx / (2l√(d²+x²)

• So total Potential Energy will be:

                 ∫ k(qdx / (2l√(d²+x²)          ( here limits of integration is from -l to l)

• Now consider cos Ф = d / √(d²+x²) .

• Differentiating, we get:

                 dx = (d / sec²Ф) dФ

• So when x --> -l, Ф --> Ф1

                 tan^-1 (-l/d)

• And when x --> l, Ф --> Ф2

                 tan^-1 (l/d)

• So therefore:

                 PE = ∫( k q cosФ d / 2l d cos²Ф ) dФ  

                 ( here limits of integration is from Ф1 to Ф2)

                 PE = ∫( k q / 2l )  sec Ф dФ  

                 ( here limits of integration is from Ф1 to Ф2)

                 PE = ( k q / 2l ) ( ln ( sec Ф + tan Ф )  )

                 Applying limit from Ф1 to Ф2

                 PE = ( k q / 2l ) ln ( √(d² + x²) + l / √(d² + x²) - l )

Answer:

         So Potential energy is ( k q / 2l ) ln ( √(d² + x²) + l / √(d² + x²) - l )