Question

There is a charge of 3.75 x 10-5 C that is placed at the corner of a cubical box with sides that are 0.50 m long. A. What is the electric flux through the sides of the box the charge is touching? B. What is the electric flux through the opposite sides of the box?

Answer 1

assume the chaarge to be at the center of a cube of side 1m . Then, the given box can be considered as on of the pieces when the 1m cube is divided into 8 equal cubes. So, flux through the opposite sides of the charge will be 1/8 th of flux through the initial 1m cube as those were the sides which were actually part of the first cube.

So, flux through the opposite 3 sides from the charge will be 1/8 th total flux

ie q/8ε₀ = 3.75 x 10^-5 /8*8.85 x10^-12 = 5.3 x10^5

As for the flux through the sides touching the charge: field lines of a charge is radially inwad or outwards so, for a plane containing a charge , the field lines will pass along its surface so,no field lines actually pass through them.

So, in this case the adjacent 3 sides are planes containing the charge so, the flux through them is zero