Question

When a \[10\mu C\] is enclosed by a closed surface, the flux passing through the surface is \[\phi \]. Now another \[10\mu C\] charge is placed inside a closed surface, then the flux passing through the surface is____.
A. \[4\phi \]
B. \[\phi \]
C. \[2\phi \]
D. Zero

Answer 1

Hint: The formula for flux passing through any surface is given by Gauss law. It is given that a \[10\mu C\] is enclosed by a closed surface and the flux passing through the surface is \[\phi \]. If two charges are placed in the closed surface then the net charge will be the addition of those two charges.

Formula Used:
According to Gauss Law, the net flux is,
\[\phi = \oint\limits_s {\overrightarrow E .\overrightarrow {dS} } = \dfrac{q}{{{\varepsilon _ \circ }}}\]

Complete step by step answer:
The total number of lines of force that pass through a closed surface in electric fields is called electric flux. It is denoted by \[\phi \]. Electric flux coming out of a small surface element \[\overrightarrow {dS} \] is given by
\[d\phi = \overrightarrow E .\overrightarrow {dS} = EdS\cos \theta \]

If \[\overrightarrow E \] is parallel to the closed surface \[\overrightarrow {dS} \], then \[\theta = {90^ \circ }\]. That is, \[\overrightarrow E \]and \[\overrightarrow {dS} \]are mutually perpendicular. Then, \[d\phi = \overrightarrow E .\overrightarrow {dS} = EdS\cos {90^ \circ } = 0\]. Therefore, \[\phi = 0\]. On the other hand, if \[\overrightarrow E \]and \[\overrightarrow {dS} \] are in the same direction, then \[\theta = {0^ \circ }\]. Thus, \[d\phi = \overrightarrow E .\overrightarrow {dS} = EdS\cos {0^ \circ } = EdS\]. Here, \[{\phi _{\max }} = ES\]. This will be the maximum value of electric flux.

According to Gauss Law, the net flux through any closed surface is equal to \[\dfrac{1}{{{\varepsilon _0}}}\] times the total electric charge enclosed by the surface. Then,
\[\phi = \oint\limits_s {\overrightarrow E .\overrightarrow {dS} } = \dfrac{q}{{{\varepsilon _ \circ }}}\]

When a \[10\mu C\] is enclosed by a closed surface, the flux passing through the surface is \[\phi \], that is \[\phi = \dfrac{{10}}{{{\varepsilon _ \circ }}}\].
And when another \[10\mu C\] charge is placed inside a closed surface, then the flux passing through the surface is \[\phi ' = \dfrac{{10 + 10}}{{{\varepsilon _ \circ }}} = \dfrac{{20}}{{{\varepsilon _ \circ }}}\].
\[\therefore \phi ' = 2\phi \]

Hence, option C is the correct answer.

Note: Electric flux entering a closed surface is taken as negative and that passing through the surface is taken as positive. In the problem, the electric flux is passing through the surface, therefore is taken as positive. The value of electric flux is independent of the distribution of charges and the separation between them inside the closed surface