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The electric field at centre of a uniformly charged hemispherical shell is E_{o} . Now 2 portions of the hemisphere are cut from either side. \alpha = \beta = \frac{\pi}{3} then E due to remaining portion at centre. 

  • Option 1)

    \frac{E_{o}}{3}

  • Option 2)

    \frac{E_{o}}{6}

  • Option 3)

    \frac{E_{o}}{2}

  • Option 4)

    0

 

Answers (1)

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As we learned

Hemispherical charged body -

At centre O 

E=\frac{\sigma }{4\epsilon _{0}}           V=\frac{\sigma R }{2\epsilon _{0}}

- wherein

 

 

The Electric field due to each part of the hemispherical surface at O is same.

E + \frac{E}{2}+ \frac{E}{2} = E_{o}

E = \frac{E_{o}}{2}

 


Option 1)

\frac{E_{o}}{3}

Option 2)

\frac{E_{o}}{6}

Option 3)

\frac{E_{o}}{2}

Option 4)

0

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Plabita

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