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A charged parallel plate capacitor is disconnected from the cell and the plates are fixed in gravity free space. A dielectric slab having same dimensions as that of the capacitor and having dielectric constant 4 is released from rest at one end. The initial force on the slab is $F_{1}$ . When the slab moves inside and fills half the space symmetrically, the force is $F_{2}$ . Find $\frac{F_{1}}{F_{2}}$

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answer is 6.25.

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Detailed Solution

$F α \frac{1}{C^{2}} ∴ \frac{F_{1}}{F_{2}} = {(\frac{C_{2}}{C_{1}})}^{2}$

$C_{1} = C, C_{2} = 2.5 C, ∴ \frac{F_{1}}{F_{2}} = 6.25$

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