Volume charge density and potential difference in sphere

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SUMMARY

The discussion focuses on calculating the potential difference between two points inside a long nonconducting cylindrical rod with a uniform charge density of 50 nC/m³. The correct potential difference, ΔV, between point A (2.0 cm from the axis) and point B (4.0 cm from the axis) is determined to be 1.7 V. The relevant equations used include E = pr/3ε₀ for the electric field and the relationship between electric field and potential difference, ΔV = Vb - Va. The application of Gauss' law is emphasized for deriving the electric field within the cylinder.

PREREQUISITES
  • Understanding of Gauss' law in electrostatics
  • Familiarity with electric field and potential difference concepts
  • Knowledge of integration techniques for electric fields
  • Basic principles of cylindrical symmetry in electrostatics
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  • Study the application of Gauss' law for cylindrical charge distributions
  • Learn how to calculate electric fields using integration techniques
  • Explore the relationship between electric field and potential difference in electrostatics
  • Review examples of potential difference calculations in nonconducting materials
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Students and educators in physics, particularly those focusing on electrostatics, as well as anyone seeking to understand electric fields and potential differences in cylindrical geometries.

fight_club_alum
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Homework Statement


The charge of uniform density 50 nC/m3 is distributed throughout the inside of a long nonconducting
cylindrical rod (radius = 5.0 cm). Determine the magnitude of the potential difference of point A (2.0 cm from the axis of the rod) and point B (4.0 cm from the axis).

a . 2.7 V

b. 2.0 V

c. 2.4 V

d. 1.7 V <--THE ANSWER

e. 3.4 V

Homework Equations


pr/3Eo = E
E * r = V

The Attempt at a Solution


E=pr/3Eo
v=ER
v=prR/3Eo
v1-v2=prR1/3Eo-prR2/3Eo
= 37.646
 
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The electric field is a function of r. So, finding the potential difference ΔV from E will require integration.
 
Last edited:
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fight_club_alum said:
pr/3Eo = E
Is this the correct formula for inside a cylinder?
 
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TSny said:
Is this the correct formula for inside a cylinder?
I don't think so but I attempted anything because I just wanted to try
May you please show and explain to me how to get the right answer because this is my first example of this kind?
Thank you
 
Try using Gauss' law to find the electric field as a function of r inside the cylinder.
 
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TSny said:
Try using Gauss' law to find the electric field as a function of r inside the cylinder.
E A = Q/eo
E (2piRah) = P * (pi * r^2 * h)/eo
E (2Ra) = P * (r^^2) / eo
E = P* (r^2) / (eo * 2 r_a)
Ea = 353
Va = 353 * 0.02 = 7.06
Vb = 7. 06
I keep getting a wrong answer
 
fight_club_alum said:
E A = Q/eo
E (2piRah) = P * (pi * r^2 * h)/eo
E (2Ra) = P * (r^^2) / eo
It is not clear what Ra stands for. But you are on the right track. Hopefully, you drew a diagram in which you have constructed your Gaussian surface. What shape did you choose for the Gaussian surface?

Ea = 353
Va = 353 * 0.02 = 7.06
This is not the correct way to get the potential at point a. You are looking for the potential difference ΔV = Vb - Va. You can find this without having to determine Vb and Va individually. You should have covered the basic relationship between ΔV and E.

See first equation here:
http://slideplayer.com/slide/903141...gral+of+electric+field+along+a+close+path.jpg
 

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