An air-filled parallel-plate capacitor has capacitance C = 8.0 x 10^-9 F. The di

Question

An air-filled parallel-plate capacitor has capacitance C = 8.0 x 10^-9 F. The distance between
its plates is 4.0 mm. The capacitor is connected to a battery that has emf 48.0 V until it is fully
charged and then it is disconnected from the battery without being discharged. After the capacitor is
disconnected from the battery, a dielectric with dielectric constant K = 3.0 is inserted between the
plates of the capacitor.

After the dielectric has been inserted, what is:
a) the potential difference between the plates of the capacitor?
b) the magnitude of the electric field at a point midway between the plates?

Explanation / Answer

Total charge Q = CV = 3.84* 10^-7 Coulomb capacitance after inserting the di-electric C'= 3*C new voltage = Q/C' = 16 V Electric field = 16/(4*10^-3) V/m = 4000 V/m (Assuming large plate which implies uniform electric field)

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