ElectrostaticsHard
Question
Two point charges $2q$ and $q$ are placed at vertex $A$ and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :
Options
A.$\frac{3q}{\varepsilon_{0}}$
B.$\frac{q}{\varepsilon_{0}}$
C.$\frac{3q}{2\varepsilon_{0}}$
D.$\frac{3q}{4\varepsilon_{0}}$
Solution
$\phi = \frac{Q_{\text{in~}}}{\varepsilon_{0}}$
$$\phi = \frac{\frac{q}{4} + \frac{q}{2}}{\varepsilon_{0}} = \frac{3q}{4\varepsilon_{0}}$$
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