Question:
A charge ' $q$ ' is placed at one corner of a cube as shown in figure. The flux of electrostatic field $E$ though the shaded area is:
Correct Option: , 3
Solution:
(3)
$\phi=\frac{q}{24 \varepsilon_{0}}$
$\phi_{T}=\left(\frac{q}{24 \varepsilon_{0}}+\frac{q}{24 \varepsilon_{0}}\right) \times \frac{1}{2}$
$\phi_{T}=\frac{q}{24 \varepsilon_{0}}$
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