Shows a uniformly charged hemisphere of radius R. It has a volume charge density $ρ$ . If the electric field at a point 2R, above its center C is E, then what is the electric field at the point 2R below its center?

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$\frac{ρ R}{6 \in_{0} + E}$

$\frac{ρ R}{12 \in_{0}} - E$

$\frac{- ρ R}{6 \in_{0} + E}$

$\frac{ρ R}{12 \in_{0} + E}$

answer is B.

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Detailed Solution

Electric field due to lower part at B is= electric field due to full sphere –electric field due to upper part $\frac{K Q}{{(2 R)}^{2}} - E = \frac{1}{4 π \in_{0}} \frac{ρ (4 / 3) π R^{3}}{4 R^{2}} - E = \frac{ρ R}{12 \in_{0}} - E .$

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