A parallel plate capacitor with an air gap has a capacitance of 70muF when the p

Question

A parallel plate capacitor with an air gap has a capacitance of 70muF when the plates are 1.4mm apart. If it stores 0.1mC on each of the plates, what is the potential difference between the plates? What is the electric field between the plates? 4. For the capacitor in the previous problem, the air Rap is replaced with water (k = 80), but the charge on the plates is kept the same. what is the electric potential difference now? What is the electric field now? 5. The capacitor in problem 3 again has its air gap replaced with water, but now the electric potential is held constant instead of the charge. What is the new amount of charge stored on the plates of the capacitor? What is the electric field between the plates?

Explanation / Answer

New capacitance will be = k C = 80 * 70 = 5600 uF

As the capacitance s increased and the potential is kept constant the charge will increase by the same factor to that of capacitor

Q = CV

Q' = 80 * .0001 = 0.008 C

Electric field is given by

E = V/d

E = 10-4/ 70*10-6*1.410-3 = 1020 N/C

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