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Help me answer: - Electrostatics - JEE Main-11

 A cone of base radius R and height h is  located in a uniform electric field \vec{E} parallel to its base. The electric flux entering the cone is :

  • Option 1)

    \frac{1}{2}EhR

  • Option 2)

    EhR

  • Option 3)

    2EhR

  • Option 4)

    4EhR

 
Answers (2)
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N neha

As we discussed in the concept

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

 

 Area of \Delta facing = \frac{1}{2}\times h\times 2R

\therefore \phi = EhR

 


Option 1)

\frac{1}{2}EhR

Option 2)

EhR

Option 3)

2EhR

Option 4)

4EhR

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