What is the surface charge density sigma?

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The discussion focuses on calculating the surface charge density (sigma) of a conducting sphere charged to 538 V, which is determined to be 3.11E-08 C/m^2. The user attempts to find the distance at which the potential is 11.0 V using the relationship between electric field, potential, and charge density. However, they encounter confusion as their calculations yield a distance smaller than the initial radius of the sphere, contradicting the expected outcome that a lower potential should correspond to a greater distance. The conversation highlights the importance of understanding the concepts of electric potential and charge distribution in solving these problems. Clarification is sought regarding the calculations and the relationship between potential and distance.
dgoudie
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[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<br /> <br /> 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
 
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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