Find the electric field at the centre of a uniformly charged semicircular ring of radius R. Linear charge density is `lamda`

Options

A. `lambda/(4pi epsilon_0 R)`
B. `lambda/(pi epsilon_0 R)`
C. `(2lambda)/(pi epsilon_0 R)`
D. `lambda/(2pi epsilon_0 R)`

Answer

D

Explanation

We consider a different elenent dl on the ring that subtends an angle `d theta` at the center of the ring, i.e. `dl=Rd theta`. So the charge on this element is `dQ=lambda Rd theta`. This element creates a field dE, which makes an angle `theta` at the center as shown in fig.

For each differential element in the upper half of the ring, there corresponds a symmetrically palced charge element in the lower half. the y-components of the field due to these symmetric elements cancle out, and the x-components remain. So we get
`dE_(x)=dE cos theta=(dQ)/(4 pi epsilon_(0)r^(2)) cos theta=(lambda(Rd theta)cos theta)/(4 pi epsilon_(0)R^(2))`
On intergrating the expression for `dE_(x)` with respect to angle `theta` in limits `theta=-(pi)//2 to theta= +pi//2` we obtain
`E=int_(-pi//2)^(+pi//2) (lambda R)/(4 pi epsilon_(0)R^(2)) cos theta d theta=(lambda)/(2 pi epsilon_(0)R)`
In terms of the total charge, say Q, on the ring `lambda=Q//pi R` and we get `E=Q//2 pi^(2) epsilon_(0)R^(2)`
If we consider the wire in the form of an are aas shown in fig. the symmetry consideration is not useful in canceling out x- and y- components of the fields , if `theta_(1)` and `theta_(2)` are different. We will intergrated `dE_(x)` as well as `dE_(y)` in limits `theta=-theta_(1)` to `theta=-theta_(2)`.

`E_(x)=int_(-theta_(1))^(+theta_(2)) (lambda R)/(4 pi epsilon__(0)R^(2)) cos theta d theta =(lambda )/(4 pi epsilon_(0)R^(2))(sin theta_(1)-sin theta_(2))`
`E_(y)=int_(-theta_(1))^(+theta_(2)) (lambda R)/(4 pi epsilon_(0)R^(2)) sin theta d theta =(lambda )/(4 pi epsilon_(0)R^(2))(cos theta_(2)+cos theta_(1))`
For a sysmmetrical are , `theta_(1)=theta_(2)`. thus `E_(y)` vanishes and
`E_(x)=(lambda sin theta)/(2 pi epsilon_(0)R)`.