The linear charge density on a dielectric ring of radius `R` vanes with `theta` as `lambda = lambda_0 cos theta//2`, where `lambda_0` is constant. Find the potential at the center `O` of the ring [in volt].

Explanation

To find the potential at the center \( O \) of a dielectric ring with a varying linear charge density, we can follow these steps:

### Step 1: Define the linear charge density
The linear charge density on the ring is given by:
\[ \lambda(\theta) = \lambda_0 \cos\left(\frac{\theta}{2}\right) \]
where \( \lambda_0 \) is a constant.

### Step 2: Express the differential charge element
The differential charge element \( dq \) on a small segment of the ring can be expressed as:
\[ dq = \lambda(\theta) \, d\ell \]
where \( d\ell = R \, d\theta \) is the arc length corresponding to the angle \( d\theta \). Thus,
\[ dq = \lambda_0 \cos\left(\frac{\theta}{2}\right) R \, d\theta \]

### Step 3: Calculate the differential potential \( dV \)
The potential \( dV \) at the center \( O \) due to the charge \( dq \) is given by:
\[ dV = k \frac{dq}{R} \]
Substituting for \( dq \):
\[ dV = k \frac{\lambda_0 \cos\left(\frac{\theta}{2}\right) R \, d\theta}{R} = k \lambda_0 \cos\left(\frac{\theta}{2}\right) \, d\theta \]

### Step 4: Integrate to find the total potential \( V \)
To find the total potential \( V \) at the center \( O \), we integrate \( dV \) from \( \theta = 0 \) to \( \theta = 2\pi \):
\[ V = \int_0^{2\pi} k \lambda_0 \cos\left(\frac{\theta}{2}\right) \, d\theta \]

### Step 5: Solve the integral
To solve the integral, we can use the substitution \( u = \frac{\theta}{2} \), which gives \( d\theta = 2 \, du \). The limits change as follows: when \( \theta = 0 \), \( u = 0 \) and when \( \theta = 2\pi \), \( u = \pi \):
\[ V = \int_0^{\pi} k \lambda_0 \cos(u) \cdot 2 \, du = 2k \lambda_0 \int_0^{\pi} \cos(u) \, du \]
The integral of \( \cos(u) \) from \( 0 \) to \( \pi \) is:
\[ \int_0^{\pi} \cos(u) \, du = [\sin(u)]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \]
Thus,
\[ V = 2k \lambda_0 \cdot 0 = 0 \]

### Final Result
The potential at the center \( O \) of the ring is:
\[ V = 0 \, \text{volts} \]