A spherically symmetric charge distribution is characterized by a charge density having the following variation : `rho (r )=rho(1-r/R) for r lt R rho (r )=0` for r`ge ` R. Where r is the distance from the centre of the charge distribution and`(rho_0) is a constant. The electric field at an internal point (r

Options

A. `(rho_0)/( in_0) (r/3 -(r^2)/(4R))`
B. `(rho_0)/( 4 in_0) (r/3 -(r^2)/(4R))`
C. `(rho_0)/( 3 in_0) (r/3 -(r^2)/(4R))`
D. `(rho_0)/( 12 in_0) (r/3 -(r^2)/(4R))`

Answer

To find the electric field at an internal point (r < R) of a spherically symmetric charge distribution characterized by the given charge density, we can follow these steps:

### Step 1: Define the Charge Density
The charge density is given by:
\[ \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \quad \text{for } r < R \]
\[ \rho(r) = 0 \quad \text{for } r \geq R \]
where \( \rho_0 \) is a constant.

### Step 2: Calculate the Charge Element
The charge element \( dq \) in a thin spherical shell of radius \( r \) and thickness \( dr \) is given by:
\[ dq = \rho(r) \cdot dV \]
The volume element \( dV \) for a spherical shell is:
\[ dV = 4\pi r^2 dr \]
Thus,
\[ dq = \rho(r) \cdot 4\pi r^2 dr = \rho_0 \left(1 - \frac{r}{R}\right) \cdot 4\pi r^2 dr \]

### Step 3: Integrate to Find Total Charge Enclosed
To find the total charge \( Q \) enclosed within a radius \( r \), we integrate \( dq \) from \( 0 \) to \( r \):
\[ Q = \int_0^r dq = \int_0^r \rho_0 \left(1 - \frac{r'}{R}\right) 4\pi (r')^2 dr' \]
Substituting \( r' \) for \( r \) in the integral:
\[ Q = 4\pi \rho_0 \int_0^r \left(1 - \frac{r'}{R}\right) (r')^2 dr' \]

### Step 4: Evaluate the Integral
Now, we can evaluate the integral:
\[ Q = 4\pi \rho_0 \left[ \int_0^r (r')^2 dr' - \frac{1}{R} \int_0^r (r')^3 dr' \right] \]
Calculating the integrals:
\[ \int_0^r (r')^2 dr' = \frac{r^3}{3}, \quad \int_0^r (r')^3 dr' = \frac{r^4}{4} \]
Substituting these results back into the equation for \( Q \):
\[ Q = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right] \]
\[ Q = \frac{4\pi \rho_0}{12} r^3 - \frac{4\pi \rho_0}{4R} r^4 \]
\[ Q = \frac{4\pi \rho_0}{12} r^3 - \frac{\pi \rho_0}{R} r^4 \]

### Step 5: Use Gauss's Law to Find Electric Field
According to Gauss's Law:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
For a spherical surface of radius \( r \):
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]
Thus,
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]

### Step 6: Substitute \( Q \) into the Electric Field Equation
Substituting the expression for \( Q \):
\[ E = \frac{1}{4\pi \epsilon_0 r^2} \left( \frac{4\pi \rho_0}{12} r^3 - \frac{\pi \rho_0}{R} r^4 \right) \]
\[ E = \frac{\rho_0}{3\epsilon_0} r - \frac{\rho_0}{4\epsilon_0 R} r^2 \]

### Final Result
Thus, the electric field at an internal point (r < R) is given by:
\[ E = \frac{\rho_0}{3\epsilon_0} r - \frac{\rho_0}{4\epsilon_0 R} r^2 \]