WBJEE · Physics · Electrostatics
Consider a region in free space bounded by the surfaces of an imaginary cube having sides of length \(a\) as shown in the figure. A charge \(+Q\) is placed at the centre \(O\) of the cube. \(P\) is such a point out side the cube that the line \(O P\) perpendicularly intersects the surface \(A B C D\) at \(R\) and also \(O R=R P=a / 2\). A charge \(+Q\) is placed at point \(P\) also. What is the total electric flux through the five faces of the cube other than \(A B C D ?\)
- A \(\frac{Q}{\varepsilon_{0}}\)
- B \(\frac{5 Q}{6 \varepsilon_{0}}\)
- C \(\frac{10 Q}{6 \varepsilon_{0}}\)
- D zero
Answer & Solution
Correct Answer
(A) \(\frac{Q}{\varepsilon_{0}}\)
Step-by-step Solution
Detailed explanation
For surface \(A B C D\) clectric flux is zero. Because at surface \(A B C D\) net electric field is zero. Using Gauss's law, \[ \oint \mathbf{E} \cdot d \mathbf{S}=\frac{Q_{\mathrm{in}}}{\varepsilon_{0}} \] Electric flux through the five faces of the cube,…
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