A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

Question

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

Tardigrade Admin · Accepted Answer

We know for a parallel plate capacitor, F=(Q2/2 A ∈ 0) since the capacitor is isolated charge will not change, hence force will remain unchanged

Q. A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

increase

decrease

remain unchanged

become zero

We know for a parallel plate capacitor,
$F = \frac{Q ^{2}}{2 A \in _{0}}$
since the capacitor is isolated charge will not change, hence force will remain unchanged

Q. A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

increase

decrease

remain unchanged

become zero

We know for a parallel plate capacitor, F=2A∈0​Q2​ since the capacitor is isolated charge will not change, hence force will remain unchanged

We know for a parallel plate capacitor,
$F = \frac{Q ^{2}}{2 A \in _{0}}$
since the capacitor is isolated charge will not change, hence force will remain unchanged