an air capacitor is given a charge of 2μC raising its potential to 200V. If on introducing a dielectric medium, its potential falls to 50V, what is the dielectric constant of the medium?

Charge, Q = 2μ C = 2×10-6 C
​Voltage, V = 200 V
Q = C0V
⇒C0=Q/V = (2×10-6)/200 = 10-8 F
Let the dielectric constant of the medium be K.
Now voltage, V = 50 V
Q = CmV
⇒Cm=Q/V = (2×10-6)/50 = 4×10-8 F
The capacitance of a capacitor with air between the plates and dielectric between the plates is related as:
Cm = KC0
⇒4×10-8 = K​×10-8
⇒K = 4

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