Question

A long string with a charge of $\lambda$ per unit length passes through an imaginary cube of edge $\ell$ . The maximum flux of electric field through the imaginary cube can be
(A) $\sqrt 3 \dfrac{{\lambda \ell }}{{{\varepsilon _o}}}$
(B) $\dfrac{{\lambda \ell }}{{{\varepsilon _o}}}$
(C) $\sqrt 2 \dfrac{{\lambda \ell }}{{{\varepsilon _o}}}$
(D) $6\dfrac{{\lambda \ell }}{{{\varepsilon _o}}}$

Answer 1

The flux passing through a closed surface having charge q inside it is given by the formula $\phi = \dfrac{q}{{{\varepsilon _o}}}$, and in the given question, the flux will be maximum if the charge inside the cube becomes maximum, which is possible if the string crosses the cube diagonally.

Complete Step by Step Answer:
The string crossing the cube has charge $\lambda$ per unit length. Hence any part of the string having length x will have charge $\lambda x$ .
Consider the cube through which this string is crossing. A cube is a closed surface and hence if the charge inside the cube is q then flux crossing the cube will be $\phi = \dfrac{q}{{{\varepsilon _o}}}$. For this flux to be maximum, the charge inside the cube should be maximum.
The maximum possible straight length inside a cube is along the body diagonal of the cube. And the length of a body diagonal of a cube is $\sqrt 3 \ell$. If the charged string is along a body diagonal then the charge inside the cube will become
$q = \lambda \left( {\sqrt 3 \ell } \right)$,
Hence the flux crossing the cube will be
$\phi = \dfrac{q}{{{\varepsilon _o}}}$,
Putting the value of q
$\phi = \dfrac{{\lambda \left( {\sqrt 3 \ell } \right)}}{{{\varepsilon _o}}}$,
Rearranging the values
$\phi = \sqrt 3 \dfrac{{\lambda \ell }}{{{\varepsilon _o}}}$
Therefore, the correct answer to the question is option : A

Note: Before attempting to solve the problem, the student needs to be able to understand Gauss Law of Electrostatics, and calculation of flux through a closed surface. Electric flux is a scalar quantity and is defined as the number of field lines passing through the closed surface. A common mistake that students make while solving such questions is not considering that the cube can have any orientation with respect to the charged string.