what is the electric flux through a cube of side a if a point charge of q is at one of its corner
Answers
As we learnt in
Gauss's Law -
Total flux linked with a closed surface called Gaussian surface.
Formula:
\phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}
- wherein
No need to be a real physical surface.
Qenc - charge enclosed by closed surface.
Eight identical cubes are required so that the given charge q appears at the centre of the bigger cube.
\therefore \phi = \frac{1}{8}\left \lfloor \frac{q}{\varepsilon _{0}} \right \rfloor =\frac{q}{8\varepsilon _{0}}
Option 1)
\frac{2q}{\varepsilon _0}
Incorrect
Option 2)
\frac{q}{8\varepsilon _0}
Correct
Option 3)
\frac{q}{\varepsilon _0}
Incorrect
Option 4)
\frac{q}{2\varepsilon _0}
Incorrect
Answer:
1/8(q/ e0)=q/8e0
Explanation:
eight identical cubes are required so that the eight charge q appears at the center of the bigger cube. There, the electric flux passing through the given cube.
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