Question

A charge $+Q$ is located somewhere inside a vertical cone such that the dept of the charge from the free surface of the cone is $H$. It is formed that the flux associated with the cone with thw curved surface is $\frac{3Q}{5{ϵ}_0}$.
If the charge is raised vertically through a height $2H$ then the flux through the curved surface is

• A. $\frac{3Q}{5{ϵ}_0}$.
• B. $\frac{2Q}{5{ϵ}_0}$.
• C. $\frac{4Q}{5{ϵ}_0}$.
• D. Zero.

Answer 1

The correct option is B $\frac{2Q}{5{ϵ}_0}$.

$\begin{array}{c}fluxthroughflatsurface=\frac{Q}{{\epsilon }_0}-\frac{3Q}{3{\epsilon }_0}\\ =\frac{23}{5{\epsilon }_0}\\ \therefore whencheisraisedto2x\\ fluxthroughcurrentsurface=\frac{2Q}{5{\epsilon }_o}\\ Hence,\\ optionBiscorrectanswer.\end{array}$