A parallel plate capacitor has its plates separated by 0.5mm as shown in the fig
ID: 2280436 • Letter: A
Question
A parallel plate capacitor has its plates separated by 0.5mm as shown in the figure. If the electric field between the plates has a magnitude of 1000 N/C
1. Find the potential (voltage) difference between the plates
2. If a proton starts at point A and moves to point B. What is the change in its potential energy? (Charge of proton -1.6x10^-19 C)
3. If the surface area of each plate is 1.0cm^2, find its capacitance, C.
I already have the correct answers to these questions I just don't know how to arrive at the answers so if you could show your work/explain how you get yours answers that would be awesome. Thanks so much!
Explanation / Answer
1) V = E*d = 1000*0.5*10^-3 = 0.5 volts
2)
delta V = -0.5 volts
delta U = q*delta V
= 1.6*10^-19*(-0.5)
= -8*10^-20 J
3) C = A*epsilon/d
= 1*10^-4*8.854*10^-12/0.5*10^-3
= 1.77*10^-12 F
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.