A half ring of radius `r` has a linear charge density `lambda`.The potential at the centre of the half ring is

Options

A. `(lambda)/(4epsioln_(0))`
B. `(lambda)/(4pi^(2)epsilon_(0)r)`
C. `(lambda)/(4piepsilon_(0)r)`
D. `(lambda)/(4piepsilon_(0)r^(2))`

Answer

To find the potential at the center of a half ring with radius \( r \) and linear charge density \( \lambda \), we can follow these steps:

### Step 1: Understanding the Charge Distribution
The half ring has a uniform linear charge density \( \lambda \). This means that for a small segment of the half ring, the charge \( dq \) can be expressed in terms of the angle \( d\theta \) as:
\[ dq = \lambda \cdot ds \]
where \( ds = r \, d\theta \) is the length of the small segment at angle \( \theta \).

### Step 2: Expressing the Charge Element
Substituting \( ds \) into the equation for \( dq \):
\[ dq = \lambda \cdot (r \, d\theta) \]

### Step 3: Calculating the Potential Contribution from \( dq \)
The potential \( dV \) at the center of the half ring due to the small charge \( dq \) is given by the formula:
\[ dV = \frac{k \, dq}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) is Coulomb's constant. Substituting \( dq \):
\[ dV = \frac{k \, (\lambda \cdot r \, d\theta)}{r} = k \lambda \, d\theta \]

### Step 4: Integrating to Find Total Potential
To find the total potential \( V \) at the center, we need to integrate \( dV \) over the angle from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):
\[ V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dV = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} k \lambda \, d\theta \]
This simplifies to:
\[ V = k \lambda \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta = k \lambda \left[ \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = k \lambda \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = k \lambda \cdot \pi \]

### Step 5: Final Expression for the Potential
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \lambda \cdot \pi \]
This simplifies to:
\[ V = \frac{\lambda}{4 \epsilon_0} \]

### Conclusion
The potential at the center of the half ring is:
\[ \boxed{\frac{\lambda}{4 \epsilon_0}} \]