A charge q is surrounded by a closed surface consisting of an inverted cone of height h and base radius R , and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is n q 6 ε 0 (in SI units). The value of n is _________.
A. A
According to the Gauss law, the total flux linked with a closed surface is 1 ε 0 times the charge enclosed by the closed surface. As we can see from the diagram, the given shape encloses the charge q , therefore from Gauss law, we can write total flux = q ε 0 . Now, the plane connecting hemispherical shape & conical shape cuts the charge in two equal parts, therefore flux through the conical part will be exactly half of the total flux value. Hence, flux through conical part is, q 2 ε 0 . Therefore, n = 3
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