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Q.

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 $\frac{{CV}^{2}}{yD}, what is the value of y$ ?

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Detailed Solution

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Electric field at the location of one of the plates (which is due to the other plate) is $E = \frac{σ}{2 ε_{0}}$

Force on this plate, $F = QE = Q \frac{σ}{2 ε_{0}} = \frac{Q^{2}}{2 {Aε}_{0}}$

But $C = \frac{{Aε}_{0}}{d}$ and $Q = CV$

$F = \frac{C^{2} V^{2}}{2 Cd} = \frac{{CV}^{2}}{2 d}$

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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 CV2yD,what is the value of y?