How Does Electric Field Influence Potential Difference and Particle Suspension?

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SUMMARY

The discussion centers on calculating potential differences and electric fields in various scenarios involving charged parallel plates. A constant electric field of 750 N/C between plates 1.5 cm apart results in a potential difference of 11.25 V. Additionally, a spark can jump between two individuals when the electric field exceeds 4.0 x 106 V/m, leading to a calculated potential difference of 6.0 x 103 V when a person is 0.15 cm away from another. An oil drop with a charge of 8.0 x 10–19 C suspended between plates with a potential difference of 1200 V experiences a weight of 1.2 x 10–13 N.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with the formula V = Ed
  • Knowledge of charge and weight calculations in physics
  • Basic concepts of electrostatics and particle suspension
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  • Study the relationship between electric field strength and potential difference in parallel plate capacitors
  • Learn about the Millikan oil drop experiment for measuring charge
  • Explore the concept of electric field strength calculations in different configurations
  • Investigate the conditions for electrical discharge and spark formation
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pharaoh
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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
 
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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.
 
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