Solution:
A charge placed at the corner of a cube is shared equally by 8 adjacent cubes. The flux through the single cube is therefore \( \Phi = \frac{Q_{total}}{8 \varepsilon_0} = \frac{8q}{8 \varepsilon_0} = \frac{q}{\varepsilon_0} \).
The total electric flux through a cube when a charge 8q is placed at one corner of the cube is :
Leave a Reply