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  • Solve this problem A parallel plate air capacitor has capacity 'C, distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel pla

A parallel plate air capacitor has capacity 'C, distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is :

  • Option 1)

    \frac{{\text{CV}^2 }} {{\text{2 d}}}

  • Option 2)

    \frac{{\text{CV}^2 }} {\text{d}}

  • Option 3)

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

  • Option 4)

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

 

Answers (1)

best_answer

As discussed

Electric Field Intensity -

vec{E}=frac{vec{F}}{q_{0}}=frac{kQ}{r^{2}}

- wherein

 

 AND

 

 

Parallel Plate Capacitor -

C=frac{epsilon _{0}A}{d}

- wherein

Area - A seperation between two plates.

 

 

Force between the pilates

F=qt \: \: \: = \: q\times \frac{d}{2\varepsilon _{o}} \: =q \times \frac{q}{2A\varepsilon _{o}}
F=\frac{q^{2}}{2\varepsilon _{o}4}

q=cv \: \: \: \Rightarrow \: \: c=\frac{\varepsilon _{o}4}{d} \: =\varepsilon _{o}4=cd

F=\frac{c^{2}v^{2}}{2cd}=\frac{cv^{2}}{2d}


Option 1)

\frac{{\text{CV}^2 }} {{\text{2 d}}}

This option is correct

Option 2)

\frac{{\text{CV}^2 }} {\text{d}}

This option is incorrect

Option 3)

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

This option is incorrect

Option 4)

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

This option is incorrect

Posted by

prateek

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