Physics · 2018
JEE Main · 15 April 2018, Shift 2 · Q14
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 solid ball of radius R has a charge density $\displaystyle \rho$ given by $\displaystyle \rho=\rho_{\mathrm{o}}(1-\mathrm{r} / \mathrm{R})$ for $\displaystyle 0 \leq \mathrm{r} \leq \mathrm{R}$. The electric field outside the ball is :
Official answer
(2)
$\displaystyle \frac{\rho_{\mathrm{o}} \mathrm{R}^3}{12 \epsilon_{\mathrm{o}} \mathrm{r}^2}$
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JEE Main 2018 Physics question, with the answer from NTA’s final answer key. Where our answers come from.