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Q.

The electric potential in a medium of dielectric constant ‘unity’ is $ϕ (x, y, z) = {ax}^{2}$ where ‘a’ is a constant of suitable dimensions. The total charge contained in a cube of dimensions $- L \leq x, y, z \leq L$ is

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a

zero

b

$- 2 a \in_{0} L^{3}$

c

$- 16 a \in_{0} L^{3}$

d

$- 4 a \in_{0} L^{3}$

answer is D.

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Detailed Solution

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$\begin{array}{l} ϕ (x, y, z) = {ax}^{2} \\ \vec{E} = - 2 ax \hat{i} \\ \int_{B} \vec{E} \cdot d \vec{s} = (2 aL \hat{i}) (L^{2} \hat{i}) = 2 {aL}^{3} \\ \int_{B}^{x = - L} \vec{E} d \vec{s} = [(- 2 a (L) \hat{i}] \cdot (L^{2} \hat{i}) = - 2 {aL}^{3} \\ \int_{}^{} \vec{E} ds = 0 (f o r a l l o t h e r f a c e s) \end{array}$

$\begin{array}{l} ∴ ∮ \vec{E} \cdot d \vec{s} = - 4 a L^{3} = \frac{q}{ϵ_{0}} \\ ∴ q = - 4 a \in_{0} L^{3} \end{array}$

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