Consider a sphere of radius R which carries a uniform charge density $ρ$ . If a sphere of radius $\frac{R}{2}$ is carved out of it, as shown, the ratio $\frac{|{\vec{E}}_{A}|}{|{\vec{E}}_{B}|}$ i.e. the magnitude of electric field ${\vec{E}}_{A} and {\vec{E}}_{B}$ , respectively, at points A and B due to the remaining portion is:

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$\frac{18}{34}$

$\frac{18}{54}$

$\frac{17}{54}$

$\frac{21}{34}$

answer is B.

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Detailed Solution

Electric field at $A (R^{'} = \frac{R}{2}) E_{A} \cdot ds = \frac{q}{ε_{0}}$
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$\Rightarrow {\vec{E}}_{A} = \frac{ρ \times \frac{4}{3} π {(\frac{R}{2})}^{3}}{ε_{0} \cdot 4 π {(\frac{R}{2})}^{2}} \Rightarrow {\vec{E}}_{A} = \frac{ρ (\frac{R}{2})}{3 ε_{0}} = (\frac{ρR}{6 ε_{0}})$
Electric fields at B:
${\vec{E}}_{B} = \frac{k \times ρ \times \frac{4}{3} π R^{3}}{R^{2}} - \frac{k \times ρ \times \frac{4}{3} π {(\frac{R}{2})}^{3}}{{(\frac{3 R}{2})}^{2}} {\vec{E}}_{B} = \frac{ρ R}{3 ε_{0}} - (\frac{1}{4 π ε_{0}}) \frac{(ρ)}{{(\frac{3 R}{2})}^{2}} \frac{4 π}{3} {(\frac{R}{2})}^{3}$
${\vec{E}}_{B} = \frac{ρR}{3 ε_{0}} - \frac{ρR}{54 ε_{0}} \Rightarrow E_{B} = \frac{17}{54} (\frac{ρR}{ε_{0}}) |\frac{E_{A}}{E_{B}}| = \frac{1 \times 54}{6 \times 17} = (\frac{9}{17}) = \frac{9}{17} \times \frac{2}{2} = \frac{18}{34}$