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Answer please! - Electrostatics - JEE Main-11

The  electric field in a region of space is  given by,   \vec{E}= E_{0}\: \hat{i}+2 E_{0}\: \hat{j} 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

 
Answers (2)
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N neha

As we discussed in

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

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

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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