A rod of length L lies along the x-axis with its left end at the origin. It has a non-uniform charge density `lamda=alphax`, where a is a positive constant.
(a) What are the units of `alpha`?
(b) Calculate the electric potential at point A where `x = - d` .

Explanation

Charge per unit length `lambda = (Q)/(L)`
Taking `O` as origin , take an element of small length `dx` at distance `x` from the origin.
Charge on element `dq = lambda dx`
Small potential at `P` due to `dq`
`dV = (1)/(4 pi in_(0)) . (dq)/((2L - x)) = (Q)/(4 pi in_(0) L) (dx)/((2L - x))`
`V_(P) = (Q)/(4 pi in_(0) L) int_(0)^(L) (dx)/((2L - x))`
`I = int_(0)^(L) (dx)/((2L - x)) = (|log_(e )(2L - x)|_(0)^(L))/(-1)`
` = - [ log_(e ) (2L - L) - log_(e ) (2L - 0)]`
`= - log_(e ) ((L)/(2L)) = -log_(e ) ((1)/(2)) = log_(e )2`
`V_(P) = (1)/(4 pi in_(0)) (Q log_(e ) 2)/(L)`
Small electric field at `P` due to `dq`
`dE = (1)/(4 pi in_(0)) . (dq)/((2L -x)^(2)) = (Q)/(4 pi in_(0)L) .(dx)/((2L - x)^(2))`
`E_(P) = (Q)/(4 pi in_(0) L) int_(0)^(L) (dx)/((2L - x)^(2))`
`I = int_(0)^(L) (dx)/((2L - x)^(2)) = int_(0)^(L) (2L - x)^(-2) dx`
`= (|(2L -x)^(-1)|_(0)^(L))/((-1) (-1)) = |(1)/((2L - x))|_(0)^(L)`
`= (1)/(2L - L) - (1)/(2L - 0) = (1)/(L) - (1)/(2L) = (1)/2L)`
`E_(P) = (1)/(4 pi in_(0)) .(Q)/(2L^(2))`