A charge `Q` is uniformly distributed only on the fourths portion of a ring of radius `R`. The elctric potential at centre of ring is

Options

A. `(3 Q)/(4 pi epsilon_(0) R)`
B. `(Q)/(4 pi epsilon_(0) R)`
C. `(3Q)/(2 pi epsilon_(0) R)`
D. `(6Q)/(4 pi epsilon_(0) R)`

Explanation

To find the electric potential at the center of a ring with a charge \( Q \) uniformly distributed over three-fourths of its circumference, we can follow these steps:

### Step 1: Understand the configuration
We have a ring of radius \( R \) with a total charge \( Q \) uniformly distributed over three-fourths of the ring. The remaining one-fourth of the ring has no charge.

### Step 2: Determine the charge distribution
Since the charge is uniformly distributed over three-fourths of the ring, we can calculate the linear charge density \( \lambda \) as follows:
\[ \lambda = \frac{Q}{\text{Length of charged portion}} = \frac{Q}{\frac{3}{4} \cdot 2\pi R} = \frac{Q}{\frac{3\pi R}{2}} = \frac{2Q}{3\pi R} \]

### Step 3: Calculate the electric potential at the center
The electric potential \( V \) at a point due to a small charge \( dq \) is given by:
\[ dV = \frac{1}{4\pi \epsilon_0} \frac{dq}{r} \]
where \( r \) is the distance from the charge to the point where the potential is being calculated. Here, \( r = R \) (the radius of the ring).

### Step 4: Integrate the potential contribution from the charged portion
The total potential \( V \) at the center of the ring can be found by integrating \( dV \) over the charged portion of the ring:
\[ V = \int dV = \int \frac{1}{4\pi \epsilon_0} \frac{dq}{R} \]
Since \( dq = \lambda d\ell \) and \( d\ell = R d\theta \) (where \( d\theta \) is the infinitesimal angle), we can express \( dq \) as:
\[ dq = \lambda R d\theta \]
Substituting this into the expression for \( V \):
\[ V = \int_0^{\frac{3\pi}{2}} \frac{1}{4\pi \epsilon_0} \frac{\lambda R d\theta}{R} = \int_0^{\frac{3\pi}{2}} \frac{\lambda}{4\pi \epsilon_0} d\theta \]
Now substituting \( \lambda = \frac{2Q}{3\pi R} \):
\[ V = \int_0^{\frac{3\pi}{2}} \frac{2Q}{3\pi R \cdot 4\pi \epsilon_0} d\theta \]
\[ V = \frac{2Q}{12\pi^2 \epsilon_0 R} \int_0^{\frac{3\pi}{2}} d\theta = \frac{2Q}{12\pi^2 \epsilon_0 R} \cdot \frac{3\pi}{2} \]
\[ V = \frac{Q}{8\epsilon_0 R} \]

### Final Result
Thus, the electric potential at the center of the ring is:
\[ V = \frac{Q}{8\epsilon_0 R} \]