Volume charge density and potential difference in sphere

In summary, volume charge density refers to the amount of electric charge per unit volume in a sphere. It is represented by the Greek letter "rho" and is measured in coulombs per cubic meter. The potential difference in a sphere is the difference in electric potential between two points, typically the center and the surface of the sphere. It is determined by the amount of charge and the distance between the two points. The potential difference can be calculated using the equation V = kQ/R, where k is the Coulomb's constant, Q is the charge, and R is the radius of the sphere. As the charge or radius changes, the potential difference also changes. Understanding these concepts is important in understanding the behavior of electric fields and the properties
  • #1
fight_club_alum
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1

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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  • #2
The electric field is a function of r. So, finding the potential difference ΔV from E will require integration.
 
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  • #3
fight_club_alum said:
pr/3Eo = E
Is this the correct formula for inside a cylinder?
 
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  • #4
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
 
  • #5
Try using Gauss' law to find the electric field as a function of r inside the cylinder.
 
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  • #6
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
 
  • #7
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
 

FAQ: Volume charge density and potential difference in sphere

1. What is volume charge density?

Volume charge density refers to the amount of electric charge per unit volume of a given material or object. It is typically denoted by the symbol ρ (rho) and is measured in units of coulombs per cubic meter (C/m^3).

2. How is volume charge density calculated for a sphere?

For a sphere, the volume charge density is equal to the total charge (Q) divided by the volume (V) of the sphere. In mathematical terms, it can be expressed as ρ = Q/V. Alternatively, it can also be calculated by multiplying the surface charge density (σ) by the radius (r) of the sphere, i.e. ρ = σr.

3. What is the relationship between volume charge density and potential difference in a sphere?

The potential difference, also known as voltage, between two points on a sphere is directly proportional to the volume charge density. In other words, as the volume charge density increases, the potential difference also increases, and vice versa.

4. How is potential difference calculated for a sphere?

The potential difference between two points on a sphere can be calculated using the formula V = kQ/r, where k is the Coulomb's constant (9x10^9 N•m^2/C^2), Q is the total charge of the sphere, and r is the distance between the two points.

5. Can the potential difference across a sphere be negative?

Yes, the potential difference across a sphere can be negative. This occurs when the charges on the sphere are of opposite signs and thus create an electric field that points inwards towards the center of the sphere. In this case, the potential at the center of the sphere will be negative relative to a reference point outside the sphere.

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