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  • Force acting upon a charged particle kept between the plates of a charged condenser is F. If one plate of the condenser is removed, then the force acting on the same particle will become

Force acting upon a charged particle kept between the plates of a charged condenser is F . If one plate of the condenser is removed, then the force acting on the same particle will become

  • Option 1)

    0

  • Option 2)

    F/2

  • Option 3)

    F

  • Option 4)

    2F

 

Answers (1)

best_answer

As we have learned

Force between Parallel Plates Capacitor -

\dpi{100} F=\frac{\sigma ^{2}A}{2\epsilon _{0}}=\frac{Q^{2}}{2\epsilon _{0}A}=\frac{CV^{2}}{2d}

 

- wherein

\sigma -Surface\: charge\: density.

 

 

  Initially F=qE and E= \frac{\sigma }{\varepsilon _0} F= \frac{q\sigma }{\varepsilon _0}

If one plate is removed, then E becomes \frac{\sigma }{2\varepsilon _0}

So F= \frac{q\sigma }{2\varepsilon }= F/2


Option 1)

0

Option 2)

F/2

Option 3)

F

Option 4)

2F

Posted by

Plabita

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