A charge ' q' is placed at one corner of a cube

Question:

A charge ' $q$ ' is placed at one corner of a cube as shown in figure.The flux of electrostatic field $\vec{E}$ through the shaded area is:

  1. $\frac{q}{4 \varepsilon_{0}}$

  2. $\frac{q}{24 \varepsilon_{0}}$

  3. $\frac{q}{48 \varepsilon_{0}}$

  4. $\frac{q}{8 \varepsilon_{0}}$


Correct Option: , 2

Solution:

flux through cube $=\frac{\mathrm{q}}{8 \epsilon_{0}}$

flux through surfaces $\mathrm{ABEH}, \mathrm{ADGH}, \mathrm{ABCD}$ will be zero

$\phi(\mathrm{EFGH})=\phi(\mathrm{DCFG})=\phi(\mathrm{EBCF})=\frac{1}{3}\left(\frac{\mathrm{q}}{8 \epsilon_{0}}\right)$

$=\frac{\mathrm{q}}{24 \epsilon_{0}}$

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