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Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Electric Charge and Field Physics Electric Charge and Field Physics Class 12th

The electric field in a region is given by $\vec{E} = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) \frac{N}{C} .$ The ratio of flux of reported field through the rectangular surface of are 0.2m² (parallel to y – z plane) to the of the surface of are 0.3m² (parallel to x – z plane) is a : b, where a = ……………

[Here $\hat{i}, \hat{j} a n d \hat{k}$ are the unit vectors along x, y and z – axes respectively]

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Vishal Baghel | Contributor-Level 10

2 weeks ago

$?_{1} = \frac{3}{5} E_{0} (0.2) N m^{2} C^{- 1}, a n d ?_{2} = \frac{4}{5} E_{0} (0.3) N m^{2} C^{- 1}$

$\Rightarrow \frac{?_{1}}{?_{2}} = \frac{3 \times 0.2 \times 5}{5 \times 0.3 \times 4} = \frac{1}{2}$

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