A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
(a) increase
(b) decrease
(c) remain unchanged
(d) become zero.

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Solution

(c) remain unchanged

The force between the plates is given by
F = $\frac{q^{2}}{2 \in_{0} A}$
Since the capacitor is isolated, the charge on the plates remains constant.
We know that the charge is conserved in an isolated system.
Thus, the force acting between the plates remains unchanged.

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A parallel plate capacitor is first charged and then isolated , and a dielectric slab is introduced between the plates. The quantity that remains unchanged is:

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A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will (a) increase (b) decrease (c) remain unchanged (d) become zero.

(c) remain unchanged The force between the plates is given by F = q22∈0A Since the capacitor is isolated, the charge on the plates remains constant. We know that the charge is conserved in an isolated system. Thus, the force acting between the plates remains unchanged.

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
(a) increase
(b) decrease
(c) remain unchanged
(d) become zero.

(c) remain unchanged

The force between the plates is given by
F = $\frac{q^{2}}{2 \in_{0} A}$
Since the capacitor is isolated, the charge on the plates remains constant.
We know that the charge is conserved in an isolated system.
Thus, the force acting between the plates remains unchanged.