A positively charged sphere of radius $r_{0}$ carries a volume charge density $ρ$ (figure). A spherical cavity of radius $r_{0} / 2$ is then scooped out and left empty. $C_{1}$ is the center of the sphere and $C_{2}$ that of the cavity. What is the direction and magnitude of the electric field at point B?

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$\frac{17 ρ r_{0}}{54 ε_{0}} l e f t$

$\frac{ρ r_{0}}{6 ε_{0}} l e f t$

$\frac{17 ρ r_{0}}{54 ε_{0}} r i g h t$

$\frac{ρ r_{0}}{6 ε_{0}} r i g h t$

answer is A.

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Detailed Solution

Electric field on surface of a uniformly charged sphere is given by $\frac{Q}{4 π ε_{0} R^{2}} = \frac{ρ R}{3 ε_{0}}$

Electric field at outside point is given by

$E = \frac{Q}{4 π ε_{0} R^{2}} = \frac{ρ R^{3}}{3 ε_{0} r^{2}}$

$E_{B} = E_{w h o l e s p h e r e} - E_{c a v i t y}$

$= \frac{ρ r_{0}}{3 ε_{0}} - \frac{ρ {(\frac{r_{0}}{2})}^{3}}{3 ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{17 ρ r_{0}}{54 ε_{0}}$

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