What is the surface charge density sigma?

In summary, the conversation discusses a problem involving a charged conducting sphere and the determination of surface charge density and potential at a specific distance. The solution involved using the equations E = sigma/epsilon_0 and E = V/r, and also finding the total charge.
  • #1
dgoudie
29
0
[SOLVED] Electric Potential

Homework Statement


20. [1pt]
A 30.6- cm-diameter conducting sphere is charged to 538 V relative to V = 0 at r =infinity. What is the surface charge density sigma?

Correct, computer gets: 3.11E-08 C/m^2 = Sigma, r=.153

21. [1pt]
At what distance will the potential due to the sphere be only 11.0 V?

Homework Equations


[tex]E=\sigma / \epsilon_{0} and

E=V/r[/tex]

The Attempt at a Solution


I already found sigma for Question number 20, and for 21 I am trying the same method, but am just solving for r.

so I got [tex]r=(V/\sigma) *\epsilon_{0} [/tex]
and everytime I do it I come out with r=0.00313 m which is smaller than the initial r, even though I know from this statement "538 V relative to V = 0 at r =infinity" that if V is smaller r must be bigger.

What am I missing? Thanks in advance
 
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  • #2
dgoudie said:
Correct, computer gets: 3.11E-08 C/m^2 = Sigma, r=.153

21. [1pt]
At what distance will the potential due to the sphere be only 11.0 V?

I'm not very sure what you mean by saying "computer gets". Do you know the concept you're applying?

The next question has nothing to do with sigma, but to do with the total charge. Find that.
 
  • #3
!

The surface charge density, sigma, is the amount of electric charge per unit area on the surface of a conductor. It is typically represented by the Greek letter sigma (σ) and is measured in coulombs per square meter (C/m^2).

To solve for the surface charge density in this problem, we can use the formula E=σ/ε0, where E is the electric field, σ is the surface charge density, and ε0 is the permittivity of free space. We can also use the formula E=V/r, where V is the potential and r is the distance from the center of the sphere.

For question 20, we are given the diameter of the sphere (30.6 cm) and the potential (538 V). We can use the formula E=σ/ε0 to solve for the surface charge density σ. Plugging in the given values, we get:

E=σ/ε0
538 V=(σ)/(8.85x10^-12 C^2/N*m^2)
σ=538 V x 8.85x10^-12 C^2/N*m^2 = 4.76x10^-9 C/m^2

For question 21, we are asked to find the distance at which the potential is 11 V. We can use the formula E=V/r to solve for r. Plugging in the given values, we get:

E=V/r
11 V=(538 V)/r
r=538 V/11 V = 48.91 cm = 0.4891 m

So, at a distance of 0.4891 m from the center of the sphere, the potential will be 11 V. This is greater than the initial distance of 30.6 cm, as expected.
 

1. What is surface charge density?

Surface charge density, denoted by the symbol sigma (σ), is a measure of the amount of electric charge present on a surface per unit area. It is a fundamental concept in the study of electricity and is used to describe the electric field at a surface.

2. How is surface charge density calculated?

Surface charge density can be calculated by dividing the total electric charge on a surface by the surface area. It is typically expressed in units of coulombs per square meter (C/m²) or coulombs per square centimeter (C/cm²).

3. What factors affect surface charge density?

The surface charge density on a surface is affected by the amount of electric charge present, the surface area, and the dielectric constant of the material. It can also be influenced by external factors such as electric fields or the presence of other charged particles.

4. How does surface charge density relate to electric field?

Surface charge density and electric field are closely related. The surface charge density is proportional to the electric field at the surface, and the direction of the electric field is always perpendicular to the surface. This relationship is described by Gauss's law.

5. Why is surface charge density important in electrostatics?

Surface charge density is an essential concept in electrostatics because it helps to describe and understand the behavior of electric fields at surfaces. It is also used in various applications, such as in the design of electronic devices and in the study of surface phenomena in materials science.

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