What is the surface charge density sigma?

[dgoudie]

[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

$$E=\sigma / \epsilon_{0} and E=V/r$$

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 $$r=(V/\sigma) *\epsilon_{0}$$
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

 

[Shooting Star]

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.

 

[Moetasim]

!

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.