A positively charged sphere of radius $r_{0}$ carries a volume charge density $ρ$ . A spherical cavity of radius $r_{0} / 2$ is then hollowed out and this portion is left empty as cavity as shown. $C_{1}$ is the centre of the sphere and $C_{2}$ is that of the cavity the direction and magnitude of the electric field at point B is

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$\frac{ρ r_{0}}{6 \in_{0}} (\hat{i})$

$\frac{ρ r_{0}}{6 \in_{0}} (- \hat{i})$

$\frac{17 ρ r_{0}}{54 \in_{0}} (\hat{i})$

$\frac{17 ρ r_{0}}{54 \in_{0}} (- \hat{i})$

answer is A.

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Detailed Solution

Electric field on surface of a uniformly charged sphere of radius R is given as
$E_{1} = \frac{Q}{4 π ε_{0} R^{3}} = \frac{ρ R}{3 ε_{0}}$
Electric field at outside point of a sphere of radius R is given as
$E_{2} = \frac{Q}{4 π ε_{0} r^{2}} = \frac{ρ R^{3}}{3 ε_{0} r^{2}}$
The electric field at point B is given as
$E_{B} = \frac{ρ r_{0}}{3 ε_{0}} = \frac{ρ {(\frac{r_{0}}{2})}^{3}}{3 ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{17 ρ r_{0}}{54 ε_{0}}$

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