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Assertion : The force between the plates of a parallel plate capacitor is proportional to charge on it.
Reason : Electric force is equal to charge per unit area.
Options:
Reason : Electric force is equal to charge per unit area.
Solution:
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Verified Answer
The correct answer is:
If both assertion and reason are false.
The potential energy stored between the plates of capacitor,
$U=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \frac{Q^2}{\varepsilon_0 A} x \quad\left(\because C=\frac{\varepsilon_0 A}{x}\right)$
Force $F=\frac{-\partial U}{\partial x}=-\frac{\partial}{\partial x}\left(\frac{1}{2} \frac{Q^2 x}{\varepsilon_0 A}\right)=-\frac{1}{2} \frac{Q^2}{\varepsilon_0 A}$
Negative sign shows force is attractive.
$U=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \frac{Q^2}{\varepsilon_0 A} x \quad\left(\because C=\frac{\varepsilon_0 A}{x}\right)$
Force $F=\frac{-\partial U}{\partial x}=-\frac{\partial}{\partial x}\left(\frac{1}{2} \frac{Q^2 x}{\varepsilon_0 A}\right)=-\frac{1}{2} \frac{Q^2}{\varepsilon_0 A}$
Negative sign shows force is attractive.
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