JEE Mains · Physics · STD 12 - 1. Electric charges and fields
The electric field in a region is given by \(\vec E = \frac{3}{5}{E_0}\hat i + \frac{4}{5}{E_0}\hat j\) and \(E_0 = 2\times10^3\, N/C\). Then, the flux of this field through a rectangular surface of area \(0.2\, m^2\) parallel to the \(y-z\) plane is......\(\frac{{N - {m^2}}}{C}\)
- A \(240\)
- B \(320\)
- C \(0\)
- D \(560\)
Answer & Solution
Correct Answer
(A) \(240\)
Step-by-step Solution
Detailed explanation
\(\phi = \overrightarrow {\rm{E}} \cdot \overrightarrow {\rm{A}} \) \(=\frac{3}{5} \mathrm{E}_{0}(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \cdot \mathrm{A} \hat{\mathrm{i}} \) \( =\frac{3}{5} \mathrm{E}_{0} \mathrm{A} \) \(=\frac{3}{5} \mathrm{E}_{0} \times 0.2 \)…
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