JEE Mains · Physics · STD 12 - 1. Electric charges and fields
The electric field in a region of space is given by, \(\overrightarrow E = {E_0}\hat i + 2{E_0}\hat j\) where \(E_0\, = 100\, N/C\). The flux of the field through a circular surface of radius \(0.02\, m\) parallel to the \(Y-Z\) plane is nearly
- A \(0.125\,Nm^2/C\)
- B \(0.02\,Nm^2/C\)
- C \(0.005\,Nm^2/C\)
- D \(3.14\,Nm^2/C\)
Answer & Solution
Correct Answer
(A) \(0.125\,Nm^2/C\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{i}}+2 \mathrm{E}_{0} \hat{\mathrm{j}}\) Given, \(\mathrm{E}_{0}=100 \,\mathrm{N} / \mathrm{c}\) So, \(\overrightarrow{\mathrm{E}}=100 \hat{\mathrm{i}}+200 \hat{\mathrm{j}}\) Radius of circular surface…
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