Let ρ(r) = Qπ R⁴r be the charge density distribution for a solid sphere of radius R and total charge Q. For a point p inside the sphere at distance r₁ from the center of the sphere, the magnitude of electric field is: (a) Q4πε₀ r₁² (b) Q r₁²4πε₀ R⁴ (c) Q r₁²3πε₀ R⁴ (d) 0

Answer: (b). Charge enclosed within r₁: q = ∫₀r₁ (Q r)/(π R⁴) 4π r² dr = (Q r₁⁴)/(R⁴). Gauss's law: E = (q)/(4πε₀ r₁²) = (Q r₁²)/(4πε₀ R⁴) So the answer is (b) (Q r₁²)/(4πε₀ R⁴).

Let ρ(r) = Qπ R⁴r be the charge density distribution for a solid… | JEE Main 2009 (AIEEE)

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Electric Flux & Gauss's Law

Let $\rho(r) = \frac{Q}{\pi R^{4}}r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$ . For a point $p$ inside the sphere at distance $r_1$ from the center of the sphere, the magnitude of electric field is:

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Answer: (b)

Charge enclosed within $r_1$ : $q = \int_0^{r_1} \frac{Q r}{\pi R^4} 4\pi r^2\,dr = \frac{Q r_1^4}{R^4}$ .
Gauss's law: $E = \frac{q}{4\pi\varepsilon_0 r_1^2} = \frac{Q r_1^2}{4\pi\varepsilon_0 R^4}$
So the answer is (b) $\frac{Q r_1^2}{4\pi\epsilon_0 R^4}$ .

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