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  • The force between the plates of a charged parallel plate capacitor is -Option: 1 <img alt="\frac{q^2}{2 \operatorname{ta} A}" src="https://entranceco

The force between the plates of a charged parallel plate capacitor is -

Option: 1

\frac{q^2}{2 \operatorname{ta} A}


Option: 2

F=\frac{q}{12 \epsilon_0 A}


Option: 3

F=\frac{q^2}{2 \epsilon_O A}


Option: 4

None of these


Answers (1)

best_answer

Capacity of parallel plate condenser = \mathrm{\frac{G_0 A}{d} }
\mathrm{\therefore }\mathrm{\therefore } Intensity of electric field between plates = \mathrm{E\, \, where \, \, E=\frac{q}{2 \operatorname{to} A}+\frac{q}{2 \operatorname{to} A}=\frac{q}{\operatorname{to} A} }
\mathrm{\therefore \quad } Electric energy density between the plates = \mathrm{U \, where, u=\frac{1}{2} \epsilon_0 E^2=\frac{1}{2} \epsilon_0 \cdot \frac{2}{\epsilon_0{ }^2 A^2}=\frac{q}{2 \epsilon_0 A^2} }
\mathrm{\therefore } force of attraction between the plates =\mathrm{F \, where, f=U \times A=\frac{q^2}{2 E_0 A^2} \times A=\frac{q^2}{2 \epsilon_0 A} }

\mathrm{F=\frac{q^2}{2 \epsilon_0 A} \text { Ans. }
 

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seema garhwal

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