A positively charged sphere of radius r_o carries a volume charge density $ρ_{E}$ . A spherical cavity of radius $r_{0} / 2$ is then scooped out and left empty, as shown. What is the direction and magnitude of the electric field at point B?

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$\frac{{ρr}_{0}}{6 ε_{0}}$ right

$\frac{17 {ρr}_{0}}{54 ε_{0}}$ right

$\frac{{ρr}_{0}}{6 ε_{0}}$ left

$\frac{17 ρ r_{0}}{54 ε_{0}} left$

answer is A.

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Detailed Solution

Electric field on surface of a uniformly charged sphere is given by
$\frac{Q}{4 {πε}_{0} R^{2}} = \frac{ρR}{3 ε_{0}}$
Electric field at outside point is given by
$\begin{array}{l} E = \frac{Q}{4 {πε}_{0} r^{2}} = \frac{{ρR}^{3}}{3 ε_{0} r^{2}} \\ {|{\bar{E}}_{r}|}_{at B} = E_{sphere} - E_{cavity} = \frac{{ρr}_{0}}{3 ε_{0}} - \frac{ρ {(\frac{r_{0}}{2})}^{3}}{3 ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{17 {ρr}_{0}}{54 ε_{0}} \end{array}$