The electric field inside a sphere having charge density related to the distance from the centre as`rho=alpha r` (`alpha` is a constant) is :

Options

A. `(alphar^(3))/(4epsilon_(0))`
B. `(alphar^(2))/(4epsilon_(0))`
C. `(alphar^(2))/(3epsilon_(0))`
D. none of these

Answer

To find the electric field inside a sphere with a charge density that varies with distance from the center, we can follow these steps:

### Step 1: Understand the Charge Density
The charge density is given as:
\[ \rho = \alpha r \]
where \(\alpha\) is a constant and \(r\) is the distance from the center of the sphere.

### Step 2: Use Gauss's Law
According to Gauss's Law, the electric flux \(\Phi\) through a closed surface is related to the charge \(Q\) enclosed by that surface:
\[ \Phi = \frac{Q}{\epsilon_0} \]
where \(\epsilon_0\) is the permittivity of free space.

### Step 3: Calculate the Charge Enclosed
To find the charge \(Q\) enclosed within a radius \(r\), we need to integrate the charge density over the volume of the sphere of radius \(r\):
\[ Q = \int \rho \, dV \]
The volume element in spherical coordinates is:
\[ dV = r'^2 \sin \theta \, dr' \, d\theta \, d\phi \]
Integrating over the volume of the sphere:
\[ Q = \int_0^r \int_0^\pi \int_0^{2\pi} \rho(r') r'^2 \sin \theta \, d\phi \, d\theta \, dr' \]
Substituting \(\rho = \alpha r'\):
\[ Q = \int_0^r \alpha r' \cdot r'^2 \, dr' \cdot \int_0^\pi \sin \theta \, d\theta \cdot \int_0^{2\pi} d\phi \]
Calculating the angular integrals:
\[ \int_0^\pi \sin \theta \, d\theta = 2, \quad \int_0^{2\pi} d\phi = 2\pi \]
Thus,
\[ Q = \alpha \cdot 2\pi \cdot \int_0^r r'^3 \, dr' \]

### Step 4: Evaluate the Integral
Now, evaluate the integral:
\[ \int_0^r r'^3 \, dr' = \frac{r^4}{4} \]
So,
\[ Q = \alpha \cdot 2\pi \cdot \frac{r^4}{4} = \frac{\alpha \pi r^4}{2} \]

### Step 5: Substitute into Gauss's Law
Now, substitute \(Q\) back into Gauss's Law:
\[ \Phi = \frac{Q}{\epsilon_0} = \frac{\frac{\alpha \pi r^4}{2}}{\epsilon_0} \]
The electric flux \(\Phi\) can also be expressed as:
\[ \Phi = E \cdot A = E \cdot 4\pi r^2 \]
where \(A\) is the surface area of the sphere.

### Step 6: Set the Two Expressions for Flux Equal
Setting the two expressions for flux equal gives:
\[ E \cdot 4\pi r^2 = \frac{\alpha \pi r^4}{2\epsilon_0} \]

### Step 7: Solve for Electric Field \(E\)
Now, solving for \(E\):
\[ E = \frac{\alpha r^4}{8\pi \epsilon_0 r^2} = \frac{\alpha r^2}{8\epsilon_0} \]

### Step 8: Final Expression
Thus, the electric field inside the sphere is:
\[ E = \frac{\alpha r^2}{8\epsilon_0} \]

### Conclusion
The electric field inside the sphere is proportional to \(r^2\) and is given by:
\[ E = \frac{\alpha r^2}{8\epsilon_0} \]