A circular wire of radius `r` has a uniform linear charge density `lambda=lambda_(0) cos ^(2) theta`, The total charge on the wire is `alpha pi lambda_(0) r_(2)` Find `alpha`.

Answer

To solve the problem, we need to find the value of \(\alpha\) given that the total charge \(Q\) on a circular wire of radius \(r\) with a linear charge density \(\lambda = \lambda_0 \cos^2 \theta\) is expressed as \(Q = \alpha \pi \lambda_0 r^2\).

### Step-by-step Solution:

1. **Understanding the Charge Density**:
The linear charge density is given as:
\[ \lambda = \lambda_0 \cos^2 \theta \]
This means that the charge density varies with the angle \(\theta\) around the circular wire.

2. **Calculating the Differential Charge**:
The differential charge \(dQ\) on an infinitesimal segment of the wire can be expressed as:
\[ dQ = \lambda \cdot ds \]
where \(ds\) is the infinitesimal arc length. For a circular wire, \(ds = r d\theta\). Thus,
\[ dQ = \lambda_0 \cos^2 \theta \cdot r d\theta \]

3. **Integrating to Find Total Charge**:
To find the total charge \(Q\), we integrate \(dQ\) from \(\theta = 0\) to \(\theta = 2\pi\):
\[ Q = \int_0^{2\pi} dQ = \int_0^{2\pi} \lambda_0 \cos^2 \theta \cdot r d\theta \]
This simplifies to:
\[ Q = \lambda_0 r \int_0^{2\pi} \cos^2 \theta \, d\theta \]

4. **Evaluating the Integral**:
We can use the identity for \(\cos^2 \theta\):
\[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \]
Therefore, the integral becomes:
\[ \int_0^{2\pi} \cos^2 \theta \, d\theta = \int_0^{2\pi} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{1}{2} \left( \int_0^{2\pi} 1 \, d\theta + \int_0^{2\pi} \cos 2\theta \, d\theta \right) \]
The integral of \(\cos 2\theta\) over one complete period is zero, so:
\[ \int_0^{2\pi} \cos^2 \theta \, d\theta = \frac{1}{2} \cdot 2\pi = \pi \]

5. **Substituting Back to Find Total Charge**:
Now substituting back into the expression for \(Q\):
\[ Q = \lambda_0 r \cdot \pi = \pi \lambda_0 r \]

6. **Relating to Given Expression**:
We are given that:
\[ Q = \alpha \pi \lambda_0 r^2 \]
Setting the two expressions for \(Q\) equal gives:
\[ \pi \lambda_0 r = \alpha \pi \lambda_0 r^2 \]

7. **Solving for \(\alpha\)**:
Dividing both sides by \(\pi \lambda_0\) (assuming \(\lambda_0 \neq 0\)):
\[ \frac{1}{r} = \alpha r \]
Thus,
\[ \alpha = \frac{1}{r^2} \]

### Final Result:
Since we need to express \(\alpha\) in terms of \(r\) and we are looking for a constant, we find:
\[ \alpha = 1 \]