How Is the Potential Calculated at the Center of a Charged Copper Sphere?

[stylez03]

Homework Statement

The electric field at the surface of a charged, solid, copper sphere is 4200 N/C, directed toward the center of the sphere. The sphere has a radius of 0.190 m.

What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Homework Equations

The potential is the same outside/insider right so that leaves:

$$V_{surfrace} = \frac {q} {4*pi*(8.85*10^{-12})*R}$$

The Attempt at a Solution

$$V_{surfrace} = \frac {4200*10^{-19}} {4*pi*(8.85*10^{-12})*0.190}$$

 

[Hootenanny]

Note that electric field is not the same as charge.

 

[stylez03]

Note that electric field is not the same as charge.

So this should involved $$\vec {E}$$ somehow right?

$$E = \frac {1} {4*pi*E_{0}} * \frac {q} {R^{2}}$$

 

[Hootenanny]

So this should involved $$\vec {E}$$ somehow right?
$$E = \frac {1} {4*pi*E_{0}} * \frac {q} {R^{2}}$$

Yes, you need to use this to determine the charge on the surface of the sphere.

 

[stylez03]

Yes, you need to use this to determine the charge on the surface of the sphere.

So the relation with Electric Field to potential would be:

$$V = q*E$$ ?

 

[Hootenanny]

So the relation with Electric Field to potential would be:

$$V = q*E$$ ?

No. Take your previous equation and use to to calculate q

 

[stylez03]

No. Take your previous equation and use to to calculate q

Ohhh I see

$$Q = \frac {R^{2}E} {K}$$

 

[Hootenanny]

Ohhh I see

$$Q = \frac {R^{2}E} {K}$$

Correct. This page may be useful http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html#c1

 

[stylez03]

Correct. This page may be useful http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html#c1

Awesome, thank you for the link references. I wish there were more solid examples in the book to base off of these questions.

 

[stylez03]

hm, I tried the following equation and it said it was wrong?

$$Q = \frac {R^{2}E} {K}$$

$$Q = \frac {(0.190)^{2} * 4200} {\frac {1} {4*pi*E_{0}}}$$

$$Q = 1.68*10^{-8}$$

 

[Hootenanny]

The question didn't ask you for the charge, it asked you for the potential inside the sphere...

 

[stylez03]

The question didn't ask you for the charge, it asked you for the potential inside the sphere...

So after finding charge q, I need to substitute it back into

$$V = \frac {Kq} {r}$$

 

[Hootenanny]

Yes...

 

[stylez03]

Yes...

I had one final question on this problem. I just got the answer, but when I did the calculation it came out to be positive value but the online program said to check my signs, meaning it should be negative. I'm not sure what the intuition is behind the solution being negative, would you know or could you explain that quickly. Thank you again for your time and patients. Starting to understand the difference for charge/potential.

 

[Hootenanny]

The clue is in the question

Homework Statement

The electric field at the surface of a charged, solid, copper sphere is 4200 N/C, directed toward the center of the sphere.

The electric field is define such that it is directed away from positive charge and towards negative charge. Or from from points of higher potential to lower potential, therefore, if the electric field is directed towards the centre of the sphere, then the centre must be at a lower potential than the outside of the sphere. Since zero potential (i.e. V=0) is taken to be at $r=\infty$, then the potential at the centre must be negative. Does that make sense? (Apologies if it doesn't but I'm running out of coffee )

Thank you again for your time and patients.

Its been a pleasure

Starting to understand the difference for charge/potential.

Once you got your head round potential vs. field vs. potential energy you'll sail the rest of electrostatics.