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` .

Options

A. `vecE=lamda/(4piepsilon_0a)(-hati)`
B. `vecE=lamda/(4piepsilon_0a)(-hati)`
C. `vecE=lamda/(2piepsilon_0a)(-hati)`
D. `vecE=lamda/(2epiepsilon_0a)(-hati)`

Answer

A

Explanation

Charge on dx is `dq = lamdadx`
Electric field due to element dx
`dE= (dq)/(4piepsilon_0x^2)implies E=int_a^(oo)(lamdadx)/(4piepsilon_0x^2)`
`implies E= (lamda)/(4piepsilon_0)[(x^(-2+1))/(-2+1)]_a^(oo) =(lamda)/(4piepsilon_0a)implies vecE =(lamda)/(4piepsilon_0a)(-hati)`