How Does Electric Field Influence Potential Difference and Particle Suspension?

  • Thread starter Thread starter pharaoh
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 6K views
pharaoh
Messages
47
Reaction score
0
A constant electric field of 750 N/C is between a set of parallel plates. What is the potential difference between the parallel plates if they are 1.5 cm apart?


V=Ed
V= (750)(0.05)= 11.25 v








7. A spark will jump between two people if the electric field exceeds 4.0 x 10^ 6 V/m.

You shuffle across a rug and a spark jumps when you put your finger 0.15 cm from another person’s arm. Calculate the potential difference between your body and the other person’s arm.

V= ED
(4.0*10^6)(0.0015)= 6.0*10^3 V









8. An oil drop having a charge of 8.0 x 10^ –19 C is suspended between two charged parallel plates. The plates are separated by a distance of 8.0 mm, and there is a potential difference of 1200 V

between the plates. What is the weight of the suspended oil drop?


V= Ed
1200/0.008= E
E= 150000



is my answers right
E= mg\q

150000= mg\ 8.0*10^-19
mg= 1.2*10^-13 N
 
Physics news on Phys.org
pharaoh said:
A constant electric field of 750 N/C is between a set of parallel plates. What is the potential difference between the parallel plates if they are 1.5 cm apart?


V=Ed
V= (750)(0.05)= 11.25 v








7. A spark will jump between two people if the electric field exceeds 4.0 x 10^ 6 V/m.

You shuffle across a rug and a spark jumps when you put your finger 0.15 cm from another person’s arm. Calculate the potential difference between your body and the other person’s arm.

V= ED
(4.0*10^6)(0.0015)= 6.0*10^3 V









8. An oil drop having a charge of 8.0 x 10^ –19 C is suspended between two charged parallel plates. The plates are separated by a distance of 8.0 mm, and there is a potential difference of 1200 V

between the plates. What is the weight of the suspended oil drop?


V= Ed
1200/0.008= E
E= 150000



is my answers right
E= mg\q

150000= mg\ 8.0*10^-19
mg= 1.2*10^-13 N
I still marvel at the ingenuity of the original experiment.
http://www68.pair.com/willisb/millikan/experiment.html
Seems like you are correct.
 
Last edited by a moderator: