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Problem 2 - Entrance Test

A parallel plate capacitor has a capacitance of 10 μF with air between the plates. If a dielectric slab of dielectric constant 5 is introduced between the plates, completely filling the space, what is the new capacitance?

Correct: C

The capacitance of a parallel plate capacitor with a dielectric is given by C' = kC, where k is the dielectric constant and C is the original capacitance. Therefore, C' = 5 * 10 μF = 50 μF.