How Does Electric Field Influence Potential Difference and Particle Suspension?

[pharaoh]

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

 

[denverdoc]

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.

 

[quicknote]

Yes, your answers for the potential difference between the parallel plates and the potential difference between your body and the other person's arm are correct. For the last question, the weight of the oil drop is not directly related to the electric field between the plates. The correct equation to use would be F=Eq, where F is the force on the charged particle, E is the electric field, and q is the charge. So the weight of the oil drop would be:

F= Eq
F= (8.0*10^-19)(1200)
F= 9.6*10^-16 N

Therefore, the weight of the suspended oil drop would be 9.6*10^-16 N.