The electric field in a region is given by E → = ( 3 5 E 0 i ^ + 4 5 E 0 j ^ ) 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 i ^ , j ^ a n d k ^ are the unit vectors along x, y and z – axes respectively]

? 1 = 3 5 E 0 ( 0 . 2 ) N m 2 C − 1 , a n d ? 2 = 4 5 E 0 ( 0 . 3 ) N m 2 C − 1 ⇒ ? 1 ? 2 = 3 * 0 . 2 * 5 5 * 0 . 3 * 4 = 1 2

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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5 months ago

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

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

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All the charge given to a conducting sphere resides on outer surface.

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