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  • Help me solve this The force between the plates of a parallel plate capacitor of capacitance Cand distance of separation of the plates dwith a potential difference Vbetween the plates, is

The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is

  • Option 1)

    \frac{CV^2}{2d}

  • Option 2)

    \frac{C^2V^2}{2d^2}

  • Option 3)

    \frac{C^2V^2}{d^2}

  • Option 4)

    \frac{V^2d}{C}

 

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.

 

 F= \frac{Q^2}{2\varepsilon _0A}\Rightarrow F= \frac{C^2V^2}{2\varepsilon _0A}= \frac{CV^2}{2d}

 


Option 1)

\frac{CV^2}{2d}

Option 2)

\frac{C^2V^2}{2d^2}

Option 3)

\frac{C^2V^2}{d^2}

Option 4)

\frac{V^2d}{C}

Posted by

Plabita

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