A solid non conducting sphere has charge density `pCm^(-3)` Electric field at distance x from its centre is ______ `[x lt R]`

Options

A. `(px)/(3epsilon_(0))`
B. `(px)/(2epsilon_(0))`
C. `(px)/(Repsilon-(0))`
D. None

Answer

To solve the problem of finding the electric field inside a solid non-conducting sphere with a uniform charge density \( \rho \) at a distance \( x \) from its center (where \( x < R \)), we can use Gauss's law. Here’s a step-by-step solution:

### Step 1: Understanding the Problem
We have a solid non-conducting sphere with a uniform charge density \( \rho \) (in \( \text{C/m}^3 \)). We need to find the electric field \( E \) at a point located at a distance \( x \) from the center of the sphere, where \( x \) is less than the radius \( R \) of the sphere.

**Hint:** Visualize the sphere and the point where you want to find the electric field.

### Step 2: Apply Gauss's Law
According to Gauss's law, the electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q_{\text{enc}} \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]

**Hint:** Remember that the electric field is uniform over the Gaussian surface if the surface is spherical.

### Step 3: Choose a Gaussian Surface
We choose a Gaussian surface that is a sphere of radius \( x \) (where \( x < R \)). The electric field \( E \) is constant over this surface, and the area \( A \) of the Gaussian surface is \( 4\pi x^2 \).

**Hint:** The symmetry of the problem allows us to use a spherical Gaussian surface.

### Step 4: Calculate the Charge Enclosed
The charge enclosed \( Q_{\text{enc}} \) within the Gaussian surface can be calculated by integrating the charge density over the volume of the sphere of radius \( x \):
\[ Q_{\text{enc}} = \int_0^x \rho \, d\tau \]
where \( d\tau = 4\pi r^2 \, dr \). Thus,
\[ Q_{\text{enc}} = \rho \int_0^x 4\pi r^2 \, dr = \rho \left[ \frac{4\pi r^3}{3} \right]_0^x = \frac{4\pi \rho x^3}{3} \]

**Hint:** Use the formula for the volume of a sphere to find the charge enclosed.

### Step 5: Substitute into Gauss's Law
Now substituting \( Q_{\text{enc}} \) into Gauss's law:
\[ E \cdot 4\pi x^2 = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{4\pi \rho x^3}{3\epsilon_0} \]

**Hint:** Keep track of the constants and ensure they are correctly substituted.

### Step 6: Solve for the Electric Field \( E \)
Now we can solve for \( E \):
\[ E \cdot 4\pi x^2 = \frac{4\pi \rho x^3}{3\epsilon_0} \]
Dividing both sides by \( 4\pi x^2 \):
\[ E = \frac{\rho x}{3\epsilon_0} \]

**Hint:** Simplify the equation carefully to isolate \( E \).

### Final Answer
The electric field at a distance \( x \) from the center of a solid non-conducting sphere (where \( x < R \)) is given by:
\[ E = \frac{\rho x}{3\epsilon_0} \]