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The  electric field in a region of space is  given by,    where E0=100 N/C. The flux of this field through a circular surface of radius 0.02 m parallel to the Y-Z plane is nearly :

• Option 1)

0.125 Nm2/C

• Option 2)

0.02 Nm2/C

• Option 3)

0.005 Nm2/C

• Option 4)

3.14 Nm2/C

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N

As we discussed in

Electric field \vec{E} through any area \vec{A} -

$\dpi{100} \phi = \vec{E}\cdot \vec{A}=EA\cos \Theta$

$\dpi{100} S.I\; unit\; -\left ( volt \right )m\; or\; \frac{N-m^{2}}{c}$

- wherein

$\underset{E}{\rightarrow}= E_0\hat{i} + 2E_0\hat{J}$

$E_0= 100W/C$

$\vec{E}= 100\hat{i} + 200 \hat{J}$

$A= \pi r^{2} = \frac{22}{7}\times 0.02\times 0.02$

$A= 1.25\times 10^{-3}\hat{i}m^{2}$

$\therefore$ New flux $\therefore\phi =E A cos\theta$

$\phi =(100\hat{i} + 200\hat{J}). 1.25\times 10^{-3}\hat{i}cos\theta$

where $\theta = 0$

$\phi = 1.25\times 10^{-3} Nm^{2}/c$

$= 0.125 Nm^{2}/C$

Option 1)

0.125 Nm2/C

Option 2)

0.02 Nm2/C

Option 3)

0.005 Nm2/C

Option 4)

3.14 Nm2/C

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