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Question

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A. $\lambda a/ \epsilon_0$

B. $\sqrt{2} \lambda a/ \epsilon_0$

C. $6 \lambda a^2/ \epsilon_0$

D. $\sqrt{3} \lambda a/ \epsilon_0$

Answer
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Charge can be obtained from charge density as:

$q=\lambda l$

Gaussâ€™s law:

"The flux through any surface enclosing charge q is $q/ \epsilon_0$ or

$\phi = \dfrac{ q}{ \epsilon_0}$

We are given a long string with line charge density $\lambda$ therefore, the charge obtained when this string passes through the cube (parallel to its base), we see that the total charge we will get will be:

$q= \lambda a$

Consider the figure:

To get maximum flux outside the box, the maximum length of this string should pass through the box. So, we choose a cross diagonal orientation of such a string. The total length of the string in the box will be given as:

$l= \sqrt{2a^2 + a^2}$

This expression comes as a result of Pythagoras theorem applied to the triangle with our body diagonal as its hypotenuse l, base diagonal $\sqrt{2} a$ as its base and one side a as its perpendicular.

This gives us total length as:

$l= \sqrt{3} a$

So, the charge becomes

$q= \lambda \sqrt{3} a$

Keeping this in Gaussâ€™s law, we get required maximum flux as:

$\phi = \dfrac{ \lambda \sqrt{3} a}{ \epsilon_0}$

From the question, the word maximum is quite important. One might naively conclude the wrong answer if the placement of the string in the box is wrong. The base diagonal $\sqrt{2} a$ is nothing but the diagonal length in a square, as the base of a cube is just a square.