Volume charge density and potential difference in sphere

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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.
 
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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
 
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