MHT CET · Physics · Magnetic Effects of Current
A charge \(q\) is placed at the corner of a cube of side \(a\). The electric flux through the cube is
- A \(\frac{q}{\varepsilon_{0}}\)
- B \(\frac{q}{3 \varepsilon_{0}}\)
- C \(\frac{q}{6 \varepsilon_{0}}\)
- D \(\frac{q}{8 \varepsilon_{0}}\)
Answer & Solution
Correct Answer
(D) \(\frac{q}{8 \varepsilon_{0}}\)
Step-by-step Solution
Detailed explanation
According to Gauss's law, the electric flux through a closed surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net charge enclosed by the surface. Since, \(q\) is the charge enclosed by the surface, then the electric flux \(\phi=\frac{q}{\varepsilon_{0}}\). If charge \(q\) is placed at a corner of cube, it will be divided into 8 such cubes. Therefore, electric flux through the cube is
\(
\phi^{\prime}=\frac{1}{8}\left(\frac{q}{\varepsilon_{0}}\right)
\)
\(
\phi^{\prime}=\frac{1}{8}\left(\frac{q}{\varepsilon_{0}}\right)
\)
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