So I got how to do part B. The new capacitance would be 2.0 F because the paper

Question

So I got how to do part B. The new capacitance would be 2.0 F because the paper is removed. Can someone explain how to solve the rest of this problem please 1. A parallel-plate capacitor with a 6.0 acitance is fully charged by a 4.0 V battery. The plates (30 points are 6.0 mm apart and have paper (K 3.0) between them. A. What is the charge on each plate? Draw the capacitor showing the key details. The battery is disconnected and the paper is removed. After equilibrium is reached again, B. What is the capacitance of the capacitor? C. What is the charge on each plate? D. What is the potential difference across the plates of the capacitor? E. What is the ratio of the electric field now to the original value when the paper was between the plates current E (E original)?

Explanation / Answer

C. Charge on capaitor remain same as earlier

Qearlier =CV=6*4=24C

D

24=2*V

=> V=12volts

E. electric field = V/d

Ecurrent/Eoriginal =Vcurrent /Voriginal = 12/4 =3

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