We are tasked with calculating the electric field inside a spherically symmetric charge distribution characterized by the charge density:
ρ(r)=ρ0(1−rR) for rwhere rr is the radial distance from the center, RR is the radius of the distribution, and ρ0\rho_0 is a constant. We need to find the electric field at a point inside (rStep 1: Total Charge Enclosed at a Radius rr
Using Gauss's law, the electric field EE at a distance rr depends on the total charge enclosed within radius rr. To calculate this, we integrate the charge density ρ(r)\rho(r) over the spherical volume:
Qenc(r)=∫0rρ(r′)⋅4πr′2 dr′,Q_{\text{enc}}(r) = \int_0^r \rho(r') \cdot 4\pi r'^2 \, dr',
where r′r' is a dummy variable for integration. Substituting the charge density:
Qenc(r)=∫0rρ0(1−r′R)4πr′2 dr′.Q_{\text{enc}}(r) = \int_0^r \rho_0 \left( 1 - \frac{r'}{R} \right) 4\pi r'^2 \, dr'.
Step 2: Split the Integral
Expand the expression:
Qenc(r)=4πρ0∫0r(r′2−r′3R)dr′.Q_{\text{enc}}(r) = 4\pi \rho_0 \int_0^r \left( r'^2 - \frac{r'^3}{R} \right) dr'.
Separate the terms:
Qenc(r)=4πρ0[∫0rr′2 dr′−1R∫0rr′3 dr′].Q_{\text{enc}}(r) = 4\pi \rho_0 \left[ \int_0^r r'^2 \, dr' - \frac{1}{R} \int_0^r r'^3 \, dr' \right].
Step 3: Evaluate Each Integral
1. First integral:
∫0rr′2 dr′=[r′33]0r=r33.\int_0^r r'^2 \, dr' = \left[ \frac{r'^3}{3} \right]_0^r = \frac{r^3}{3}.
2. Second integral:
∫0rr′3 dr′=[r′44]0r=r44.\int_0^r r'^3 \, dr' = \left[ \frac{r'^4}{4} \right]_0^r = \frac{r^4}{4}.
Substitute these results back:
Qenc(r)=4πρ0[r33−1R⋅r44].Q_{\text{enc}}(r) = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right].
Simplify:
Qenc(r)=4πρ0(r33−r44R).Q_{\text{enc}}(r) = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right).
Step 4: Use Gauss's Law
From Gauss's law, the electric field E(r)E(r) at a distance rr is:
E(r)=14πε0⋅Qenc(r)r2.E(r) = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q_{\text{enc}}(r)}{r^2}.
Substitute Qenc(r)Q_{\text{enc}}(r):
E(r)=14πε0⋅4πρ0(r33−r44R)r2.E(r) = \frac{1}{4\pi \varepsilon_0} \cdot \frac{4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{r^2}.
Simplify:
E(r)=ρ0ε0(r3−r24R).E(r) = \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right).
The electric field at an internal point rE(r)=ρ0ε0(r3−r24R).E(r) = \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right).
The correct option is:
B) ρ0ε0(r3−r24R)\frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right).
