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The electric field in a certain region is acting radially outward is given by $E = A r$ . A charge contained in a sphere of radius ‘a’ centred at the origin of the field, will be given by

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$4 π \in_{0} A a^{2}$

$A \in_{0} a^{2}$

$4 π \in_{0} A a^{3}$

$\in_{0} A a^{3}$

answer is B.

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Detailed Solution

$ϕ = (A r) (4 π r^{2})$

$r = a$

$ϕ = (A a) (4 π a^{2})$

$\frac{q_{i n}}{\in_{0}} = A 4 π a^{3} \Rightarrow q_{i n} = 4 π \in_{0} a^{3} A$

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