JEE Mains · Physics · STD 12 - 1. Electric charges and fields
In a cuboid of dimension \(2 L \times 2 L \times L\), a charge \(q\) is placed at the centre of the surface ' \(S\) ' having area of \(4 L ^2\). The flux through the opposite surface to ' \(S\) ' is given by
- A \(\frac{ q }{12 \varepsilon_0}\)
- B \(\frac{ q }{3 \varepsilon_0}\)
- C \(\frac{ q }{2 \varepsilon_0}\)
- D \(\frac{q}{6 \varepsilon_0}\)
Answer & Solution
Correct Answer
(D) \(\frac{q}{6 \varepsilon_0}\)
Step-by-step Solution
Detailed explanation
\(\phi=\frac{ Q / \varepsilon_0}{6}\) Flux passing through shaded face \(=\frac{ q }{6 \varepsilon_0}\)
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