JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Medium
- Numerical
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......\(\frac{{N - {m^2}}}{C}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(240\)
Step-by-step Solution
Detailed explanation
\(\phi = \overrightarrow {\rm{E}} \cdot \overrightarrow {\rm{A}} \)
\(=\frac{3}{5} \mathrm{E}_{0}(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \cdot \mathrm{A} \hat{\mathrm{i}} \)
\( =\frac{3}{5} \mathrm{E}_{0} \mathrm{A} \)
\(=\frac{3}{5} \mathrm{E}_{0} \times 0.2 \)
\(=\frac{3}{5} \times 2 \times 10^{3} \times 0.2 \)
\( \phi =240 \frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{C}} \)
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