Let there be a spherically symmetric charge distribution with charge density varying as `rho(r)=rho(5/4-r/R)` upto `r=R`, and `rho(r)=0` for `rgtR`, where r is the distance from the origin. The electric field at a distance r(rltR) from the origin is given by

Options

A. `(4pi rho_(0)r)/(3epsilon_(0))((5)/(3)-(r)/(R))`
B. `(rho_(0)r)/(4epsilon_(0))((5)/(3)-(r)/(R))`
C. `(4rho_(0)r)/(3epsilon_(0))((5)/(4)-(r)/(R))`
D. `(rho_(0)r)/(3epsilon_(0))((5)/(4)-(r)/(R))`

Answer

To find the electric field at a distance \( r \) (where \( r < R \)) from the origin for a spherically symmetric charge distribution with the given charge density, we can follow these steps:

### Step 1: Define the charge density
The charge density is given as:
\[ \rho(r) = \rho \left( \frac{5}{4} - \frac{r}{R} \right) \quad \text{for } r < R \]
and \( \rho(r) = 0 \) for \( r \geq R \).

### Step 2: Calculate the volume of a thin spherical shell
Consider a thin spherical shell of radius \( r \) and thickness \( dr \). The volume \( dV \) of this shell is given by:
\[ dV = 4\pi r^2 dr \]

### Step 3: Calculate the charge in the thin shell
The charge \( dq \) in the thin shell can be expressed as:
\[ dq = \rho(r) \cdot dV = \rho \left( \frac{5}{4} - \frac{r}{R} \right) \cdot 4\pi r^2 dr \]

### Step 4: Calculate the total charge enclosed within radius \( r \)
To find the total charge \( Q_{\text{enclosed}} \) within a radius \( r \), we integrate \( dq \) from \( 0 \) to \( r \):
\[ Q_{\text{enclosed}} = \int_0^r dq = \int_0^r \rho \left( \frac{5}{4} - \frac{r'}{R} \right) 4\pi (r')^2 dr' \]
Here, \( r' \) is a dummy variable for integration.

### Step 5: Perform the integration
Substituting \( dq \) into the integral:
\[ Q_{\text{enclosed}} = 4\pi \rho \int_0^r \left( \frac{5}{4} (r')^2 - \frac{r'}{R} (r')^2 \right) dr' \]
This simplifies to:
\[ Q_{\text{enclosed}} = 4\pi \rho \left[ \frac{5}{4} \cdot \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right] \]
\[ = \frac{4\pi \rho}{12} \cdot 5 r^3 - \frac{4\pi \rho}{4R} r^4 \]
\[ = \frac{5\pi \rho r^3}{3} - \frac{\pi \rho r^4}{R} \]

### Step 6: Apply Gauss's Law
According to Gauss's Law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
The left-hand side becomes:
\[ E \cdot 4\pi r^2 \]
Thus, we have:
\[ E \cdot 4\pi r^2 = \frac{1}{\epsilon_0} \left( \frac{5\pi \rho r^3}{3} - \frac{\pi \rho r^4}{R} \right) \]

### Step 7: Solve for the electric field \( E \)
Rearranging gives:
\[ E = \frac{1}{4\pi \epsilon_0 r^2} \left( \frac{5\pi \rho r^3}{3} - \frac{\pi \rho r^4}{R} \right) \]
\[ = \frac{\rho}{4\epsilon_0} \left( \frac{5r}{3} - \frac{r^2}{R} \right) \]

### Final Expression
Thus, the electric field at a distance \( r \) from the origin is:
\[ E = \frac{\rho}{4\epsilon_0} \left( \frac{5r}{3} - \frac{r^2}{R} \right) \]