Electric Potential (charged conducting sphere)

In summary, the conversation discusses the calculation of surface charge density for a 33-cm-diameter conducting sphere charged to 457 V. The surface charge density is found by dividing the electric potential by the surface area of the sphere. The conversation also touches on the conversion of units from Coulombs to microCoulombs, and the importance of working in appropriate units from the beginning.
  • #1
GDGirl
50
0

Homework Statement


A 33 -cm-diameter conducting sphere is charged to 457 V (relative to a point an infinite distance from the sphere where the potential is zero).

a. What is the surface charge density σ?



Homework Equations


Surface Area= 4(pi)r^2
Surface Charge Density= Q/A
V=kQ/r



The Attempt at a Solution


So initially I found the surface area (it's 3421.1944 cm^2) and then divided the electric potential by that. This is obviously wrong.
Then, I used the formula for electric potential and solved for Q to get the charge of the sphere. I got 8.388x10^-9 C. I think I may have done something wrong because my units didn't cancel out quite right, but I didn't think too much about it. I then plugged this into my formula for surface charge density, and got 2.452x10^-10 C/m^2. My answer is supposed to be in uC/m^2 so I multiplied the answer by 10^6 and got 2.452x10^-4 uC/m^2.
I put this into the answer box, and it is wrong.
So thenI thought maybe I plugged in the wrong distance before (I used the radius of the sphere and not the diameter though I'm sure the radius is correct) and did the same as above, and ended up with 4.412x10^-3 uC/m^2.
So right now, I'm simply at a loss as for what I'm doing wrong.

I'd like to note that I'm in an algebra-based physics class for a reason- calculus is lost on me, I'm sorry.
 
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  • #2
First of all maybe express your total charge in μC.

V = kQ/r

Q = V*r/k = 457*.165/9*109 = 8.37*10-9C = .008378 μC
 
  • #3
LowlyPion said:
First of all maybe express your total charge in μC.

V = kQ/r

Q = V*r/k = 457*.165/9*109 = 8.37*10-9C = .008378 μC
that's what I meant by 'uC'. I'm not sure how to make the symbol on my keyboard.

Did I change units wrong? to go from Coloumbs to μC you would multiply by 106, right?

EDIT: I just tried that and I come up with the same answer.
 
Last edited:
  • #4
GDGirl said:
that's what I meant by 'uC'. I'm not sure how to make the symbol on my keyboard.

Did I change units wrong? to go from Coloumbs to μC you would multiply by 106, right?

Only if you are simultaneously changing to μC which are 10-6C

So you can convert 10-6 to μ, yes.
 
  • #5
GDGirl said:
that's what I meant by 'uC'. I'm not sure how to make the symbol on my keyboard.

Did I change units wrong? to go from Coloumbs to μC you would multiply by 106, right?

EDIT: I just tried that and I come up with the same answer.

I get a different answer when I divide .008378 by the surface area .3421 m2
 
  • #6
LowlyPion said:
I get a different answer when I divide .008378 by the surface area .3421 m2

Wait- .3421 m2? Wouldn't 3421.1944 cm2 be 34.211944 m2? This may be where my arithmetic is messing up.
 
  • #7
Ah- I got it now, I was converting my cm2 to m2 wrong. I've got the answer now, thank you!
 
  • #8
GDGirl said:
Ah- I got it now, I was converting my cm2 to m2 wrong. I've got the answer now, thank you!

Maybe it's better to work in m from the beginning?

Good luck.
 
  • #9
LowlyPion said:
Maybe it's better to work in m from the beginning?

Good luck.

This is a good idea. It's easy to forget for me, but I'll have to keep it in mind.
Thanks again.
 

Related to Electric Potential (charged conducting sphere)

1. What is electric potential?

Electric potential is the measure of the amount of electrical potential energy that is present at a specific point in space. It is defined as the amount of work needed to move a unit positive charge from an infinitely faraway point to the specific point in question.

2. How is electric potential different from electric field?

Electric potential is a scalar quantity whereas electric field is a vector quantity. Electric potential only depends on the electric charge and its distance from the point in question, while electric field also takes into account the direction of the force on a charge. Additionally, electric potential is measured in volts while electric field is measured in newtons per coulomb.

3. What is the electric potential of a charged conducting sphere?

The electric potential of a charged conducting sphere depends on the distance from the center of the sphere. At a distance r from the center, the electric potential is given by V = kQ/r, where k is the Coulomb constant and Q is the charge on the sphere.

4. How is the electric potential of a charged conducting sphere affected by the charge and radius?

The electric potential of a charged conducting sphere is directly proportional to the charge on the sphere and inversely proportional to the radius. This means that as the charge increases, the electric potential also increases, while as the radius increases, the electric potential decreases.

5. How does the electric potential of a charged conducting sphere vary with distance?

The electric potential of a charged conducting sphere follows an inverse relationship with distance. This means that as the distance from the center of the sphere increases, the electric potential decreases. This relationship can be mathematically expressed as V ∝ 1/r, where V is the electric potential and r is the distance from the center of the sphere.

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