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. (a) `(rho_0r)/(4epsilon_0)(5/3-r/R)`
B. (b) `(4pirho_0r)/(3epsilon_0)(5/3-r/R)`
C. (c) `(4rho_0r)/(4epsilon_0)(5/4-r/R)`
D. (d) `(rho_0r)/(3epsilon_0)(5/4-r/R)`

Explanation

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: Understand the Charge Density
The charge density is given by:
\[ \rho(r) = \rho \left( \frac{5}{4} - \frac{r}{R} \right) \quad \text{for } r < R \]
and \( \rho(r) = 0 \) for \( r > R \).

### Step 2: Use Gauss's Law
According to Gauss's Law, the electric field \( E \) at a distance \( r \) from the center of a spherically symmetric charge distribution can be found using:
\[ \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the charge enclosed by a Gaussian surface of radius \( r \).

### Step 3: Calculate the Area of the Gaussian Surface
The area \( A \) of the Gaussian surface (a sphere of radius \( r \)) is:
\[ A = 4\pi r^2 \]

### Step 4: Calculate the Enclosed Charge \( Q_{\text{enc}} \)
To find \( Q_{\text{enc}} \), we need to integrate the charge density over the volume of the sphere of radius \( r \):
\[ Q_{\text{enc}} = \int_0^r \rho(r') dV \]
where \( dV = 4\pi r'^2 dr' \). Thus,
\[ Q_{\text{enc}} = \int_0^r \rho \left( \frac{5}{4} - \frac{r'}{R} \right) 4\pi r'^2 dr' \]

### Step 5: Substitute the Charge Density
Substituting the expression for \( \rho(r') \):
\[ Q_{\text{enc}} = 4\pi \rho \int_0^r \left( \frac{5}{4} - \frac{r'}{R} \right) r'^2 dr' \]

### Step 6: Evaluate the Integral
Now, we evaluate the integral:
\[ Q_{\text{enc}} = 4\pi \rho \left( \int_0^r \frac{5}{4} r'^2 dr' - \int_0^r \frac{r'}{R} r'^2 dr' \right) \]
Calculating each integral:
1. \(\int_0^r r'^2 dr' = \frac{r^3}{3}\)
2. \(\int_0^r r'^3 dr' = \frac{r^4}{4}\)

Thus,
\[ Q_{\text{enc}} = 4\pi \rho \left( \frac{5}{4} \cdot \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right) \]
Simplifying this gives:
\[ Q_{\text{enc}} = 4\pi \rho \left( \frac{5r^3}{12} - \frac{r^4}{4R} \right) \]

### Step 7: Substitute \( Q_{\text{enc}} \) into Gauss's Law
Now substituting \( Q_{\text{enc}} \) into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{4\pi \rho \left( \frac{5r^3}{12} - \frac{r^4}{4R} \right)}{\epsilon_0} \]
Thus,
\[ E = \frac{\rho \left( \frac{5r^3}{12} - \frac{r^4}{4R} \right)}{r^2 \epsilon_0} \]

### Step 8: Simplify the Expression for \( E \)
Simplifying the expression for \( E \):
\[ E = \frac{\rho}{\epsilon_0} \left( \frac{5r}{12} - \frac{r^2}{4R} \right) \]

### Final Answer
Thus, the electric field at a distance \( r \) from the origin (for \( r < R \)) is given by:
\[ E = \frac{\rho}{\epsilon_0} \left( \frac{5r}{12} - \frac{r^2}{4R} \right) \]