Consider a parallel plate capacitor with a capacitance of 2.2x10^-6 F . Initiall

Question

Consider a parallel plate capacitor with a capacitance of 2.2x10^-6 F. Initially, there is no net electric charge on each plates. When a certain amount of electric charge is transfered from one plate to the other, the potential difference between two plates becomes 1.5 V.

a) The proton is released from the positively charged plate and travels to the negatively charged plate. What is its kinetic energy when it arrives at the negatively charged plate?

b) Assuming that the distance between the plates is 5.1 mm, what is the magnitude of the force exerted on the proton?

Details please!

Explanation / Answer

kinetic energy=qV=1.6*10^-19*1.5=2.4*10^-19J=1.5eV

electric field=V/d=1.5/0.0051=294.117N/C

force=qE=1.6*10^-19*294.117=4.7*10^-17N

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