A circle of radius a has charge density given by `lambda=lambda_(0)cos^(2)theta` on its circumference, where `lambda_(0)` is a positive constant and `theta` is the angular position of a point on the circle with respect to some reference line. The potential at the centre of the circle is

Options

A. `(lambda_(0))/(4epsilon_(0))`
B. zero
C. `(lambda_(0))/(2epsilon_(0))`
D. `(lambda_(0))/(epsilon_(0))`

Answer

To find the potential at the center of a circle of radius \( a \) with a charge density given by \( \lambda = \lambda_0 \cos^2 \theta \) on its circumference, we can follow these steps:

### Step 1: Understand the Charge Distribution
The charge density \( \lambda \) varies with the angle \( \theta \) around the circle. The total charge on an infinitesimal arc length \( dx \) at angle \( \theta \) can be expressed as:
\[ dq = \lambda \, dx \]
where \( dx \) is the length of the arc corresponding to the angle \( d\theta \).

### Step 2: Relate Arc Length to Angle
The arc length \( dx \) can be expressed in terms of the radius \( a \) and the angle \( d\theta \):
\[ dx = a \, d\theta \]
Thus, the infinitesimal charge becomes:
\[ dq = \lambda \, a \, d\theta = \lambda_0 \cos^2 \theta \cdot a \, d\theta \]

### Step 3: Write the Expression for Potential
The potential \( V \) at the center of the circle due to a charge \( dq \) at a distance \( a \) is given by:
\[ dV = \frac{1}{4 \pi \epsilon_0} \frac{dq}{a} \]
Substituting for \( dq \):
\[ dV = \frac{1}{4 \pi \epsilon_0} \frac{\lambda_0 \cos^2 \theta \cdot a \, d\theta}{a} = \frac{\lambda_0 \cos^2 \theta}{4 \pi \epsilon_0} d\theta \]

### Step 4: Integrate to Find Total Potential
To find the total potential \( V \) at the center, we integrate \( dV \) from \( \theta = 0 \) to \( \theta = 2\pi \):
\[ V = \int_0^{2\pi} dV = \int_0^{2\pi} \frac{\lambda_0 \cos^2 \theta}{4 \pi \epsilon_0} d\theta \]
This simplifies to:
\[ V = \frac{\lambda_0}{4 \pi \epsilon_0} \int_0^{2\pi} \cos^2 \theta \, d\theta \]

### Step 5: Evaluate the Integral
Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \):
\[ \int_0^{2\pi} \cos^2 \theta \, d\theta = \int_0^{2\pi} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{1}{2} \left( \int_0^{2\pi} 1 \, d\theta + \int_0^{2\pi} \cos 2\theta \, d\theta \right) \]
The integral of \( 1 \) over \( [0, 2\pi] \) is \( 2\pi \) and the integral of \( \cos 2\theta \) over one full period is \( 0 \):
\[ \int_0^{2\pi} \cos^2 \theta \, d\theta = \frac{1}{2} (2\pi + 0) = \pi \]

### Step 6: Substitute Back to Find Potential
Substituting back into the expression for \( V \):
\[ V = \frac{\lambda_0}{4 \pi \epsilon_0} \cdot \pi = \frac{\lambda_0}{4 \epsilon_0} \]

### Final Answer
The potential at the center of the circle is:
\[ V = \frac{\lambda_0}{4 \epsilon_0} \]