A solid ball of radius R has a charge density ρ given by ρ = ρ o (1 -r/R) for 0 ≤ r ≤ R. The electric field outside the ball is: (a) ρo R³εo r² (b) ρo R³12εo r² (c) 4ρo R³3εo r² (d) 3ρo R³4εo r²

Answer: (b). Total charge: Q = ∫₀R ρ₀(1 - r/R)4π r² dr = 4πρ₀(R³/3 - R³/4) = (πρ₀ R³)/(3) Outside: E = (Q)/(4πε₀ r²) = (ρ₀ R³)/(12ε₀ r²) So the answer is (b).

A solid ball of radius R has a charge density ρ given by ρ = ρ o (1… | JEE Main 2018 (15 Apr, Shift 2)

Q108JEE Main 2018 · 15 APR · SHIFT 2Difficult

Electric Flux & Gauss's Law

A solid ball of radius R has a charge density $\rho$

given by $\rho = \rho$ o (1 $-\frac{r}{R}$ ) for 0 $\leq$ r $\leq$ R. The electric field outside the ball is:

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Answer: (b)

Total charge: $Q = \int_0^R \rho_0\left(1 - \frac{r}{R}\right)4\pi r^2 dr = 4\pi\rho_0\left(\frac{R^3}{3} - \frac{R^3}{4}\right) = \frac{\pi\rho_0 R^3}{3}$
Outside: $E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{\rho_0 R^3}{12\varepsilon_0 r^2}$
So the answer is (b).

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Electric Flux & Gauss's Law question from JEE Main 2018 (15 Apr, Shift 2)

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