A parallel plate capacitor having its lower end fixed and upper end is attached with spring having spring constant K. Upper plate is in equilibrium before switch is closed. After switch is closed, the condition on the potential of battery so that the system can acquire new equilibrium position is $V \leq \sqrt{{(\frac{p}{q})}^{r} \frac{L^{3} K}{ε_{0} A}}$ . Then $p + q + r$ is? [p & q are smallest possible integers and V is potential difference across the battery ]

Question

A parallel plate capacitor having its lower end fixed and upper end is attached with spring having spring constant K. Upper plate is in equilibrium before switch is closed. After switch is closed, the condition on the potential of battery so that the system can acquire new equilibrium position is $V \leq \sqrt{{(\frac{p}{q})}^{r} \frac{L^{3} K}{ε_{0} A}}$ . Then $p + q + r$ is? [p & q are smallest possible integers and V is potential difference across the battery ]

Infinity Learn · Accepted Answer

Let in new equilibrium the upper plate is displaced by x units. If we get x less than L then only new equilibrium is achieved : Kx=12A∈0[A∈0L−xV]2 Kx(L−x)2=A∈0V22 Maximum value of Kx(L−x)2=? K(L−x)2+2Kx(L−x)(−1)=0 x=L/3 Maximum value of Kx(L−x)2=K4L327So for real x K4L227≥A∈02V2 ⇒V≤827L3KA∈0 ⇒V≤(23)3L3KA∈0 or maximum kinetic energy of string in its first overtone a02π2Tl

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