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: Define the Charge Element
The charge density on the circumference of the circle is given by:
\[ \lambda = \lambda_0 \cos^2 \theta \]
The length of a small arc element \( dL \) at an angle \( \theta \) is given by:
\[ dL = a \, d\theta \]
Thus, the small charge \( dq \) on this arc element can be expressed as:
\[ dq = \lambda \, dL = \lambda_0 \cos^2 \theta \cdot a \, d\theta \]

### Step 2: Calculate the Potential Contribution
The potential \( dV \) at the center of the circle due to the small charge \( dq \) is given by:
\[ dV = k \frac{dq}{r} \]
where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the center of the circle, which is equal to \( a \). Substituting \( dq \) into the equation gives:
\[ dV = k \frac{\lambda_0 a \cos^2 \theta \, d\theta}{a} = k \lambda_0 \cos^2 \theta \, d\theta \]

### Step 3: Integrate to Find Total Potential
To find the total potential \( V \) at the center, we integrate \( dV \) from \( \theta = 0 \) to \( \theta = 2\pi \):
\[ V = \int_0^{2\pi} dV = \int_0^{2\pi} k \lambda_0 \cos^2 \theta \, d\theta \]
Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \), we can rewrite the integral:
\[ V = k \lambda_0 \int_0^{2\pi} \frac{1 + \cos 2\theta}{2} \, d\theta \]
This simplifies to:
\[ V = \frac{k \lambda_0}{2} \left( \int_0^{2\pi} 1 \, d\theta + \int_0^{2\pi} \cos 2\theta \, d\theta \right) \]

### Step 4: Evaluate the Integrals
The first integral evaluates to:
\[ \int_0^{2\pi} 1 \, d\theta = 2\pi \]
The second integral evaluates to:
\[ \int_0^{2\pi} \cos 2\theta \, d\theta = 0 \]
Thus, we have:
\[ V = \frac{k \lambda_0}{2} (2\pi + 0) = k \lambda_0 \pi \]

### Step 5: Final Expression
Therefore, the potential at the center of the circle is:
\[ V = k \lambda_0 \pi \]
We can express \( k \) in terms of \( \epsilon_0 \) as \( k = \frac{1}{4\pi \epsilon_0} \):
\[ V = \frac{\lambda_0 \pi}{4 \pi \epsilon_0} = \frac{\lambda_0}{4 \epsilon_0} \]

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