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  • Need clarity, kindly explain! . Force of attraction between the plates of a parallel plate capacitor is

.  Force of attraction between the plates of a parallel plate capacitor is

  • Option 1)

    \frac{q^2}{2\varepsilon _0AK}

  • Option 2)

    \frac{q^2}{\varepsilon _0AK}

  • Option 3)

    \frac{q}{2\varepsilon _0A}

  • Option 4)

    \frac{q^2}{2\varepsilon _0A^2K}

 

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.

 

 

Force on one plate due to another is

F = qE = q*\frac{\sigma }{2\varepsilon _0K}= q\left ( \frac{q}{2AK\varepsilon _0} \right )= \frac{q^2}{2AK\varepsilon _0} 

(where  \frac{\sigma }{2\varepsilon _0K}  is the electric field produced by one plate at the location of other).

 

 


Option 1)

\frac{q^2}{2\varepsilon _0AK}

Option 2)

\frac{q^2}{\varepsilon _0AK}

Option 3)

\frac{q}{2\varepsilon _0A}

Option 4)

\frac{q^2}{2\varepsilon _0A^2K}

Posted by

Plabita

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