Let `P(r)=(Q)/(piR^4)r` be the charge density distribution for a solid sphere of radius R and total charge Q. For a point 'p' inside the sphere at distance `r_1` from the centre of the sphere, the magnitude of electric field is:

Options

A. (a) `(Q)/(4piin_0r_1^2)`
B. (b) `(Qr_1^2)/(4pi in_0R^2)`
C. (c) `(Qr_1^2)/(3pi in_0R^4)`
D. (d) 0

Explanation

To find the magnitude of the electric field at a point \( P \) inside a solid sphere of radius \( R \) and total charge \( Q \) with a given charge density distribution \( P(r) = \frac{Q}{\pi R^4} r \), we can follow these steps:

### Step 1: Understand the Charge Density
The charge density \( \rho \) is given by:
\[ \rho(r) = \frac{Q}{\pi R^4} r \]
This indicates that the charge density increases linearly with the distance \( r \) from the center of the sphere.

### Step 2: Use Gauss's Law
To find the electric field at a distance \( r_1 \) from the center of the sphere, we will use Gauss's Law, which states:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux through a Gaussian surface and \( Q_{\text{enc}} \) is the charge enclosed by that surface.

### Step 3: Define the Gaussian Surface
Consider a Gaussian surface in the shape of a sphere of radius \( r_1 \) (where \( r_1 < R \)). The electric field \( E \) is uniform on this surface and directed radially outward. The area of the Gaussian surface is:
\[ A = 4\pi r_1^2 \]
Thus, the electric flux through this surface is:
\[ \Phi_E = E \cdot A = E \cdot 4\pi r_1^2 \]

### Step 4: Calculate the Enclosed Charge
Now, we need to find the charge enclosed \( Q_{\text{enc}} \) within the Gaussian surface of radius \( r_1 \):
\[ Q_{\text{enc}} = \int_0^{r_1} \rho(r) \, dV \]
where \( dV = 4\pi r^2 dr \). Substituting the expression for \( \rho(r) \):
\[ Q_{\text{enc}} = \int_0^{r_1} \left(\frac{Q}{\pi R^4} r\right) (4\pi r^2) \, dr \]
\[ = \frac{4Q}{R^4} \int_0^{r_1} r^3 \, dr \]
Calculating the integral:
\[ \int_0^{r_1} r^3 \, dr = \left[\frac{r^4}{4}\right]_0^{r_1} = \frac{r_1^4}{4} \]
Thus,
\[ Q_{\text{enc}} = \frac{4Q}{R^4} \cdot \frac{r_1^4}{4} = \frac{Q r_1^4}{R^4} \]

### Step 5: Apply Gauss's Law
Now substituting \( Q_{\text{enc}} \) back into Gauss's Law:
\[ E \cdot 4\pi r_1^2 = \frac{Q r_1^4}{\epsilon_0 R^4} \]
Solving for \( E \):
\[ E = \frac{Q r_1^4}{4\pi \epsilon_0 R^4 r_1^2} \]
\[ E = \frac{Q r_1^2}{4\pi \epsilon_0 R^4} \]

### Final Result
Thus, the magnitude of the electric field at a point \( P \) inside the sphere at a distance \( r_1 \) from the center is:
\[ E = \frac{Q r_1^2}{4\pi \epsilon_0 R^4} \]