A quarter ring of radius `R` is having uniform charge density `lambda`. Find the electric field and potential at the centre of the ring.

Explanation

To take element on circular length , take a small length of angle `d theta` at angle `theta` as shown.
`dl = R d theta`
`dq = lambda dl = lambda R d theta`
`dE_(0) = (1)/(4 pi in_(0)).(dq)/(R^(2)) = (lambda d theta)/(4 pi in_(0) R)`
This expression cannot be integrated because direction of electric field at `O` is different for various small `dq`.The components of `dE_(0)` along `x`- and `y`-axis can be integrated.
`E_(x) = int dE_(x) = int dE cos theta = (lambda)/(4 pi in_(0) R) int_(-pi//2)^(pi//2) cos theta d theta`
`int_(-pi//2)^(pi//2) cos theta d theta = |sin theta|_(-pi//2)^(pi//2) = sin(pi//2) - sin (-pi//2)`
`= 1 - (-1) = 2`
`E_(x) = (lambda)/(2 pi in_(0) R)`
`E_(y) = int dE_(y) = int dE sin theta = (lambda)/(4 pi in_(0) R) int_(-pi//2)^(pi//2) sin theta d theta`
`int_(-pi//2)^(pi//2) sin theta d theta = | - cos theta|_(-pi//2)^(pi//2) = 0`
`E_(y) = 0` or we can say by symmetry , `E_(y) = 0`
Electric field at `O`
`E_(0) = E_(x) = (lambda)/(4 pi in_(0) R)`
`V_(0) = (1)/(4 pi in_(0)) .(lambda . pi R)/(R) = (lambda)/(2 in_(0))`