Question #319996

 

2. The strength of the uniform electric field between two parallel conducting plates separated by 4.00 cm is 7.50×104 V/m . a) What is the potential difference between the plates?  b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?  


1
Expert's answer
2022-03-29T12:28:08-0400

(a) the potential difference

V=Ed=7.5104V/m×0.04m=3000  VV=Ed\\ =7.5*10^4\:\rm V/m\times0.04\: m=3000\; V

(b) the potential 1.00 cm from zero potential plate

V1=Ed1=7.5104V/m×0.01m=750  VV_1=Ed_1\\ =7.5*10^4\:\rm V/m\times0.01\: m=750\; V


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS