Search any question & find its solution
Question:
Answered & Verified by Expert
A hollow spherical shell of radius $r$ has a uniform charge density $\sigma$. It is kept in a cube of edge $3 r$ such that the centres of the cube and the shell coincide. Then the electric flux coming out of one face of a cube is ( $\varepsilon_0$ - permittivity of free space)
Options:
Solution:
2203 Upvotes
Verified Answer
The correct answer is:
$\frac{2 \pi r^2 \sigma}{3 \varepsilon_0}$
Charge on the hollow sphere, $q=\sigma \times 4 \pi r^2$
According to Gauss's law, The flux through a single face
of the cube, $\oint^{\prime}=\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}=\frac{\sigma \times 4 \pi \mathrm{r}^2}{6 \varepsilon_0}=\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}$
According to Gauss's law, The flux through a single face
of the cube, $\oint^{\prime}=\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}=\frac{\sigma \times 4 \pi \mathrm{r}^2}{6 \varepsilon_0}=\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.