Question

A spherically symmetric charge distribution is considered with charge density varying as
\(ρ(r) =   \begin{cases}     ρ_0(\frac 34−\frac rR)      &; \text{for} \ r≤R\\     Zero  &; \text{for}\ r>R    \end{cases}\)
Where, r(r<R) is the distance from the centre O (as shown in figure) The electric field at point P will be :

• A. \(\frac {ρ_0r}{4ε_0}(\frac 34−\frac rR)\)
• B. \(\frac {ρ_0r}{3ε_0}(\frac 34−\frac rR)\)
• C. \(\frac {ρ_0r}{4ε_0}(1−\frac rR)\)
• D. \(\frac {ρ_0r}{5ε_0}(1−\frac rR)\)

Answer 1

The Correct Option is C

The problem involves calculating the electric field at a point \( P \) within a spherically symmetric charge distribution where the charge density varies with radius as given: 

\(ρ(r) = \begin{cases} ρ_0(\frac{3}{4} - \frac{r}{R}) & ; \, r \leq R \\ 0 & ; \, r > R \end{cases}\)

To determine the electric field at a point \( P \), we apply Gauss's law which relates the electric flux through a closed surface to the charge enclosed by that surface.

Step-by-Step Solution

1. Consider a Gaussian sphere of radius \( r \) (where \( r \leq R \)) centered at \( O \), the center of the charge distribution.
2. The charge enclosed by this Gaussian surface can be found by integrating the charge density over the volume of the sphere: 
   \(Q_{\text{enc}} = \int_0^r ρ(r') \cdot 4 \pi {r'}^2 \, dr'\)
3. Substitute the expression for \( ρ(r) \): 
   \(Q_{\text{enc}} = \int_0^r ρ_0 \left(\frac{3}{4} - \frac{r'}{R}\right) 4 \pi {r'}^2 \, dr'\)
4. Simplify the integral: 
   \(Q_{\text{enc}} = 4 \pi ρ_0 \left[\left(\frac{3}{4} \int_0^r {r'}^2 \, dr'\right) - \left(\frac{1}{R} \int_0^r {r'}^3 \, dr'\right)\right]\)
5. Calculate the integrals: 
   \(\int_0^r {r'}^2 \, dr' = \frac{r^3}{3}\) 
   \(\int_0^r {r'}^3 \, dr' = \frac{r^4}{4}\)
6. Substitute back into the expression for \( Q_{\text{enc}} \): 
   \(Q_{\text{enc}} = 4 \pi ρ_0 \left[ \frac{3}{4} \cdot \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right]\) 
   \(Q_{\text{enc}} = π ρ_0 \left(r^3 - \frac{r^4}{R}\right)\)
7. Apply Gauss's law: 
   \(\Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{ε_0}\) 
   Since the electric field \( \mathbf{E} \) is radial and constant over the Gaussian surface, 
   \(E \cdot 4 \pi r^2 = \frac{π ρ_0 (r^3 - \frac{r^4}{R})}{ε_0}\)
8. Solve for \( E \): 
   \(E = \frac{ρ_0 r}{4 ε_0} \left(1 - \frac{r}{R}\right)\)

Thus, the electric field at point \( P \) inside the charge distribution is \(\frac {ρ_0r}{4ε_0}(1−\frac rR)\), which matches the correct answer.

This solution uses symmetry and integration over spherical volumes to determine the charge enclosed, effectively applying Gauss's law.