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(C) When a dielectric slab is placed near the plates of a charged capacitor, it experiences an attractive force directed into the region between the p

Vedclass · Accepted Answer

It is pulled into the space between the plates and will remain there.

Vedclass · Answer

(C) When a dielectric slab is placed near the plates of a charged capacitor, it experiences an attractive force directed into the region between the plates. This is because the electric field of the capacitor induces polarization charges on the dielectric, creating an attractive interaction. The potential energy of the system $U = \frac{Q^2}{2C}$ decreases as the dielectric enters the capacitor because the capacitance $C$ increases. Since every physical system tends to move towards a state of minimum potential energy, the dielectric slab is pulled into the region between the plates. Once it is fully inside, the force becomes zero (assuming a uniform field), and it remains there in equilibrium. Therefore, the correct option is $(c)$.

$A$ frictionless dielectric plate $S$ is placed on a frictionless table $T$. $A$ charged parallel plate capacitor $C$ (whose plates are also frictionless) is kept near it. The plate $S$ is partially between the plates of the capacitor. When the plate $S$ is released, what happens?

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