# Electric potential at the origin

1. Jul 7, 2009

### physics10189

1. The problem statement, all variables and given/known data
A sphere with radius 65 cm has its center at
the origin. Equal charges of 4 μC are placed at
60◦ intervals along the equator of the sphere.

What is the electric potential at the origin?

2. Relevant equations

V=KQ/R

3. The attempt at a solution

I just plug the numbers in this equation but no good.
I think it might have to do with the degree sign or its on the origin.
Note: I did not do sine or cosine for the degree because my notes did not show any sine or cosine for electric potential.

2. Jul 7, 2009

### rl.bhat

I just plug the numbers in this equation but no good.
Will you show this calculation?
In the interval of 60 degrees, how many charges you can place on the equator?

3. Jul 7, 2009

### physics10189

V=(9*10^9)(4*10^-6)/(.65)

however i looked at cramster and it said

V=6(9*10^9)(4*10^-6)/(.65)
why is that?
is it because charges can be at 10 degree difference so 60/10 = 6 charges?

4. Jul 7, 2009

### rl.bhat

No. 360 degrees/60 degrees. Total angle subtended by the radius at the center of the circle is 360 degrees.

5. Jul 7, 2009

### physics10189

If this is true about 360/60
What is the electric potential at the north pole?

So 360/90= 4 however this is not true in cramster

I use 90 degree because that is the angle from the origin to north

but cramster do not use that

they use the same degree 60 and change r to

0.65m*sqrt(2m)

why is it sqrt(2) and why 6 still?

6. Jul 7, 2009

### rl.bhat

Equal charges of 4 μC are placed at
60◦ intervals along the equator of the sphere.

The charges are placed at the interval of 60 degrees.
What is the electric potential at the north pole?
They have asked the potential at the center only.

0.65m*sqrt(2m) And from where you got this term?

7. Jul 7, 2009

### physics10189

So what does Interval mean??

The north pole question is another question I would like to ask

and cramster got that new r value for the solution

8. Jul 7, 2009

### rl.bhat

You have state the complete problem.
If you divide the circle into six equal part, each part will make an angle sixty degrees at the center. Therefore there are six charges on the equator.When you want to find the potential at the pole, the distance between the equator and the pole is 0.65*sqrt2 m

9. Jul 7, 2009

### physics10189

Well how do you know the distance is .65 sqrt(2) m?

10. Jul 7, 2009

### ideasrule

According to the superposition principle, potentials add arithmetically. That means if you have 2 charges and charge 1 would create a potential P1 in isolation, and charge 2 would create P2 in isolation, the total potential would be P1+P2. The superposition principle applies no matter how many charges you have; in this case, you have 6, and due to symmetry they all create the same potential. All you have to do is find the potential created by one of the charges and multiply the result by six.

11. Jul 7, 2009

### ideasrule

It's calculated, with the help of a diagram. You know the distance from the center to the sphere's surface, and you know the distance from the center to the pole, so what's the straight-line distance from the surface to the pole?

12. Jul 8, 2009

### physics10189

Well I have no clue what you are talking about the first reply
could you explain it better?

I believe this could he a 30 60 triangle

but its sides are 1:2:sqrt(3)

could you tell me why it is sqrt (2) instead???

13. Jul 8, 2009

### rl.bhat

In a sphere distance from the center to the equator and distance from center to the pole is equal and it is equal to R ( radius of the sphere). These distances are perpendicular to each other. Hence distance between any point on the equator to the pole is d = ( R^2 + R^2)^1/2