Question

A charge q is placed at one corner of a cube. The electric flux through any of the three faces adjacent to the charge is zero. The flux through any one of the other three faces is
A. $\dfrac{q}{{3{ \in _0}}}$
B. $\dfrac{q}{{6{ \in _0}}}$
C. $\dfrac{q}{{12{ \in _0}}}$
D. $\dfrac{q}{{24{ \in _0}}}$

Answer 1

Hint: Use Gauss Law
Charge is placed at a corner of a cube so that it can be visualised as being at the centre of a cube of double its side i.e.,comprising 8 such cubes kept beside and top of each other.

Formula used: The total flux from a charge q is $\dfrac{q}{{{ \in _0}}}$

Complete step by step solution:
According to the Gauss Law, the total flux from a charge q is q/ε₀.
If the charge is on the corner of a cube, some of the flux enters the cube and leaves through some of its faces. But some of the flux doesn't enter the cube.
Suppose the charge is on a corner of each of the 8 cubes. An equal amount of flux will spread outwards symmetry through each of the cubes.
Charge for one cube = $\dfrac{q}{8}$
Now in any given cube it is touching 3 of its faces. So, the area vector of that side and the electric field vector will be perpendicular. So, flux through those 3 sides will be 0. (Field and area would be perpendicular so the dot product is zero).
Equal amount of flux will flow from the other 3 sides.
So, flux through one side = $\dfrac{{\left( {\dfrac{q}{8}} \right)}}{3}*\dfrac{1}{{{ \in _0}}}$=$\dfrac{q}{{24{ \in _0}}}$

Correct answer is : D. $\dfrac{q}{{24{ \in _0}}}$

Note: Here is a brief idea of few terms used in the solution:
Vector is a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another.
Dot product is used for defining lengths and angles.
Area vector is the vector of a plane surface whose magnitude is the area of the figure and whose direction is that of a perpendicular to the plane of the figure.