Question

A total charge q is distributed uniformly along a straight rod of length l the potential at point p

Answer 1

see figure, seperation between straight rod and point P is h.

cut an elementary length of straight rod of thickness dx at x distance from the midpoint of it.

so, charge on element, dq = (q/L)dx

potential at point P , V = Kdq/√(x² + h²)

= K(q/L)$\int\limits^{L/2}_{-L/2}{\frac{1}{\sqrt{x^2+h^2}}}\,dx$

we know, ∫dx/(x² + a²) = ln(x + √(x² + a²) + C

so, V = k(q/L)$\left[ln(x+\sqrt{x^2+h^2})\right]^{L/2}_{-L/2}$

= k(q/L)[ln{L/2 + √(L² + 4h²)/2 } - ln{-L/2 + √(L² + 4h²)/2}]

= kq/L ln[ {L + √(L² + 4h²)}/{-L+√(L² + 4h²)}]

Answer 2

$\huge\bold\purple{Answer:-}$

A total charge Q is distributed uniformly along a straight rod of Length L. The potential at a point P at a distance h from the midpoint of the rod is (Hint: ∫x2+a2 1dx=ln(x+x2+a2 ))