A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is $ρ$ . The centre of sphere and centre of cavity are separated by a distance a. The magnitude of electric field inside the cavity is

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$E = \frac{ρa}{3 ε_{0}}$

$E = \frac{ρa}{ε_{0}}$

$E = \frac{ρ r}{3 ε_{0}}$

$E = \frac{ρ r}{ε_{0}}$

answer is A.

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Detailed Solution

${\vec{E}}_{1} = \frac{ρ \vec{r}}{3 ε_{0}}$

${\vec{E}}_{2} = \frac{- ρ (\vec{r} - \vec{a})}{3 ε_{0}}$

$\vec{E} = {\vec{E}}_{1} + {\vec{E}}_{2} = \frac{ρ \vec{r}}{3 ε_{0}} + [\frac{- ρ (\vec{r} - \vec{a})}{3 ε_{0}}] = + \frac{ρ \vec{a}}{3 ε_{0}} = constant.$

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