A line charge of length frac{a}{2} is kept at | vedclass

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(A) The total length of the line charge is $L = \frac{a}{2}$.The total charge $q$ of the line charge is $q = \lambda L = \lambda \left( \frac{a}{2} \r

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$\frac{\lambda a}{8 \epsilon_0}$

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(A) The total length of the line charge is $L = \frac{a}{2}$.The total charge $q$ of the line charge is $q = \lambda L = \lambda \left( \frac{a}{2} \right) = \frac{\lambda a}{2}$.This line charge is placed at the center of an edge of the cube. An edge is shared by $4$ identical cubes in a symmetric arrangement.Therefore, the fraction of the charge enclosed by the given cube is $\frac{1}{4}$ of the total charge.Thus, the charge enclosed by the cube is $q_{in} = \frac{q}{4} = \frac{\lambda a / 2}{4} = \frac{\lambda a}{8}$.According to Gauss's Law, the total electric flux $\phi$ through the cube is $\phi = \frac{q_{in}}{\varepsilon_0}$.Substituting the value of $q_{in}$, we get $\phi = \frac{\lambda a}{8 \varepsilon_0}$.

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