JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A point charge of \(+\,12 \,\mu C\) is at a distance \(6 \,cm\) vertically above the centre of a square of side \(12\, cm\) as shown in figure. The magnitude of the electric flux through the square will be ....... \(\times 10^{3} \,Nm ^{2} / C\)
- A \(452\)
- B \(381\)
- C \(226\)
- D \(113\)
Answer & Solution
Correct Answer
(C) \(226\)
Step-by-step Solution
Detailed explanation
From symmetry \(\phi=\frac{1}{6}\left(\frac{ q }{\varepsilon_{0}}\right)\) \(=\frac{12 \times 10^{-6}}{6 \times 8.85 \times 10^{-12}}\) \(=225.98 \times 10^{3} \,\frac{ Nm ^{2}}{ s }\) \(\simeq 226 \times 10^{3} \, \frac{ Nm ^{2}}{ C }\)
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