# Potential of Charged Sphere

• stylez03
In summary, the electric field at the surface of a charged, solid, copper sphere is 4200 N/C, directed towards the center of the sphere. The sphere has a radius of 0.190 m.
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

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

Note that electric field is not the same as charge.

Hootenanny said:
Note that electric field is not the same as charge.

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

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

stylez03 said:
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.

Hootenanny said:
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:

$$V = q*E$$ ?

stylez03 said:
So the relation with Electric Field to potential would be:

$$V = q*E$$ ?
No. Take your previous equation and use to to calculate q

Hootenanny said:
No. Take your previous equation and use to to calculate q

Ohhh I see

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

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

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

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

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

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

Hootenanny said:
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

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

Yes...

Hootenanny said:
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.

The clue is in the question
stylez03 said:

## 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 :grumpy:)
stylez03 said:
Thank you again for your time and patients.
Its been a pleasure
stylez03 said:
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.

## 1. What is the potential of a charged sphere?

The potential of a charged sphere is the amount of work required to move a unit positive charge from infinity to a point on the surface of the sphere, measured in volts (V).

## 2. How is the potential of a charged sphere calculated?

The potential of a charged sphere can be calculated using the equation V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the charge on the sphere, and r is the radius of the sphere.

## 3. What is the relationship between potential and distance from a charged sphere?

The potential of a charged sphere decreases as the distance from the sphere increases. This relationship follows an inverse square law, meaning that the potential decreases by a factor of four as the distance doubles.

## 4. How does the potential of a charged sphere vary with its charge?

The potential of a charged sphere is directly proportional to its charge. This means that as the charge on the sphere increases, the potential also increases.

## 5. What is the significance of the potential of a charged sphere?

The potential of a charged sphere is an important concept in understanding the behavior of electric fields. It helps to determine the direction and strength of the electric field, as well as the energy associated with the charged sphere.

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