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Question

(A) The electric field is given by $\overrightarrow{E} = E_0 \hat{i} + 2E_0 \hat{j}$.Given $E_0 = 100 \, N/C$,we have $\overrightarrow{E} = 100 \hat{i

Vedclass · Accepted Answer

$0.125 \, Nm^2/C$

Vedclass · Answer

(A) The electric field is given by $\overrightarrow{E} = E_0 \hat{i} + 2E_0 \hat{j}$.Given $E_0 = 100 \, N/C$,we have $\overrightarrow{E} = 100 \hat{i} + 200 \hat{j} \, N/C$.The circular surface is parallel to the $Y-Z$ plane,so its area vector $\overrightarrow{A}$ points in the $\hat{i}$ direction.The area $A = \pi r^2 = \pi 	imes (0.02)^2 = 3.14159 	imes 0.0004 \approx 1.256 	imes 10^{-3} \, m^2$.Thus,$\overrightarrow{A} = 1.256 	imes 10^{-3} \hat{i} \, m^2$.The electric flux $\phi$ is given by $\phi = \overrightarrow{E} \cdot \overrightarrow{A}$.$\phi = (100 \hat{i} + 200 \hat{j}) \cdot (1.256 	imes 10^{-3} \hat{i})$.Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = 0$,we get $\phi = 100 	imes 1.256 	imes 10^{-3} = 0.1256 \, Nm^2/C$.Rounding to the nearest option,the flux is approximately $0.125 \, Nm^2/C$.

The electric field in a region of space is given by $\overrightarrow E = E_0 \hat i + 2E_0 \hat j$,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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