The electric field in a region of space is gi | vedclass

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Question

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{i}}+2 \mathrm{E}_{0} \hat{\mathrm{j}}$

Given, $\mathrm{E}_{0}=100 \,\mathrm{N} / \mathrm{c

Vedclass · Accepted Answer

$0.125\,Nm^2/C$

Vedclass · Answer

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{i}}+2 \mathrm{E}_{0} \hat{\mathrm{j}}$

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 $=0.02\, \mathrm{m}$

Area $=\pi r^{2}=\frac{22}{7} 	imes 0.02 	imes 0.02$

$=1.25 	imes 10^{-3} \hat{\mathrm{i}}\, \mathrm{m}^{2}$ [Loop is parallel to $\mathrm{Y}-\mathrm{Z}$ plane ]

Now, flux $(\phi)=$ $EA \,cos$ $	heta$

$=(100 \hat{\mathrm{i}}+200 \hat{\mathrm{j}}) \cdot 1.25 	imes 10^{-3} \mathrm{i} \cos 	heta^{\circ}\left[	heta=0^{\circ}ight]$

$=125 	imes 10^{-3}\, \mathrm{Nm}^{2} / \mathrm{c}$

$=0.125\, \mathrm{Nm}^{2} / \mathrm{c}$

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

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