Question

The electric field at the center of a uniformly charged ring is zero. What is the electric field at the center of a half-ring if the charge on it be Q its radius be R?
(A) ${\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{\pi R^2}}$
(B) ${\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{R^2}}$
(C) ${\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{2Q}{\pi R^2}}$
(D)${\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{2Q}{R^2}}$

Answer 1

Hint: The electric field at the center of a uniformly charged ring is zero but there will be some electric field at the center of a half ring.

Complete step-by-step answer:
We are interested to find out the electric field at the center of the half ring. Let us calculate the electric field E at a point P located along the x-axis of a uniformly charged half ring. Let λ represent the charge distribution per unit length along the half-ring then we have the following.
$\lambda ={\displaystyle \frac{dq}{ds}}$ -(1)

Let us consider the above half ring. Let the point at which we are interested to find out the electric field.
We know that The net charge on the length of the circumference of the half-ring is given as below,
Q=λ × ${\displaystyle \frac{\pi }{2}}$ R-(2)
Where λ is equal to the charge distribution, R is equal to the distance of point charge from the center of the ring.
We know the force due to the electric field is given as follows,
$E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q}{R^2}}$ -(3)
Put values of Q in the equation of electric field from equation (1) in (3)
And on differentiation to find out the small element dE we get,
On differentiation we get,
$dE={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{\lambda ds}{R^2}}$ -(4)
We know that, due to symmetry, the horizontal components get to cancel out and therefore we need to calculate the electric field along with the vertical component i.e. y-axis only.we can be as follows,
i.e.$\text{d}{\text{E}}_{\text{y}}=dEsin\theta$-(5)
Substitute equation (4) in (5) we get,
For the entire electric field on the ring, sum all the small elements dE as follows,
$E_y=\mathrm{\Sigma }{\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{\lambda ds}{R^2}}\sin \theta$
Lets Substitute equation (2), we get
$E_y=\mathrm{\Sigma }{\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{\lambda \pi R}{2R^2}}\sin \theta$
On canceling R from denominator and numerator, we get,
$E_y=\mathrm{\Sigma }{\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{\lambda \pi }{2R^{}}}\sin \theta$
Lets, Integrate the equation from 0 to ${\displaystyle \frac{\pi }{2}}$, we get
$E_y={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{2Q}{\pi R^2}}\underset{0}{\overset{{\displaystyle \frac{\pi }{2}}}{\int }}\sin \theta d\theta$
⟹$E_y={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{2Q}{\pi R^2}}$
Therefore, the electric field at the center of half-ring is ${\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{2Q}{\pi R^2}}$
Hence, option (C) is the correct option.

Additional information:
The electrical field is a vector quantity, it has both magnitude and direction .electrical field on the half-ring is canceled out by the electrical field produced at the other half ring. As the electrical field in half, the ring is in the opposite direction to that in the other half ring.

Note:
The electric field is defined as the force per unit charge.
There are two types of electric fields
(1) static electric field
(2) dynamic or varying electric field.
The strength of the electric field is defined by the electric field lines.
In a uniform electric field, field lines are straight, parallel, and uniformly spaced.
If the electric field is zero that means electric field lines do not exist in that region.