Question

Which one of the following figures represents the radial electric field distribution E_R caused by a spherical cloud of electrons with a volume charge density,

ρ = -3ρ₀ for 0 ≤ R ≤ a (both ρ₀, a are positive and R is the radial distance),

and ρ = 0 for R > a?

• A. Fig. (i)
• B. Fig. (ii)
• C. Fig. (iii)
• D. Fig. (iv)

Hint:

For spherically symmetric charge distributions, use Gauss's Law. The electric field inside varies linearly with \( R \), and outside it decays as \( 1/R^2 \).

Answer 1

The Correct Option is C

We use Gauss’s Law for spherical symmetry: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{{enc}}}{\varepsilon_0} \] The enclosed charge for radius \( R \leq a \) is: \[ Q_{{enc}} = \int_0^R (-3\rho_0) \cdot 4\pi r^2 \, dr = -4\pi \rho_0 R^3 \] Thus, \[ E_R(R) \cdot 4\pi R^2 = \frac{-4\pi \rho_0 R^3}{\varepsilon_0} \quad \Rightarrow \quad E_R(R) = \frac{-\rho_0 R}{\varepsilon_0} \] For \( R>a \), the total charge enclosed is: \[ Q_{{enc}} = -3\rho_0 \cdot \frac{4}{3}\pi a^3 = -4\pi \rho_0 a^3 \quad \Rightarrow \quad E_R(R) = \frac{-\rho_0 a^3}{\varepsilon_0 R^2} \] Hence: \( E_R \) increases in magnitude (negatively) linearly inside the sphere (for \( R<a \)). \( E_R \) decays as \( 1/R^2 \) outside the sphere (for \( R>a \)). 
This matches Fig. (iii).