# Potnetial of a spherical Shell

• stunner5000pt
In summary, the conversation discusses problem 3.28 in Griffith's textbook, which involves calculating the dipole moment of a spherical shell with a surface charge dependent on theta. The conversation considers equ. 3.98, which uses vectors for p and r', and concludes that the dipole moment cannot be zero.
stunner5000pt
Griffith's EM problem 3.28
A spherical shell of radius R has a surface charge $\sigma = k \cos \theta$

a) Calculate the dipole moment of this charge distribution.
i know that
$$p = r' \sigma(r') da'$$

but here sigma depends on theta
would the dipole moment p then turn into
$$p = \theta' \sigma(theta') da'$$

and the radius of the sphere is constant theta and phi are constant
so that
$$p = \int_{0}^{\pi} \int_{0}^{2 pi} \theta' \sigma(\theta') R^2 \sin\theta' d \theta' d \phi$$
i get a negative dipole moemnt as a result of this though... which amkes no sense
what am i doing wrong??

thanks :)

Last edited:
Look at equ. 3.98. p and r' are VECTORS in there.

I take it you are referring to Griffith's textbook?

quasar987 said:
Look at equ. 3.98. p and r' are VECTORS in there.

right they are vectors...

so then i can't use theta the way i used it

so
$$\vec{p} = \int \vec{r'} \simga(\theta') d\vec{a'}$$
$$p = \int_{0}^{\pi} \int_{0}^{2\pi} r' k \cos\theta' r'^2 \sin\theta d\theta d\phi$$

but the integral
$$\int_{0}^{2\pi} \cos\theta' \sin\theta' d\theta = 0$$!
cant have zero dipole moment...

HalfManHalfAmazing said:
I take it you are referring to Griffith's textbook?

problem 3.28
page 151

$$\vec{r}=r\hat{r}=r(\hat{x}\sin\theta \cos \phi+\hat{y}\sin\theta\sin\phi+\hat{z}\cos\theta)$$

$\hat{r}$ is not a vector like $\hat{x},\hat{y},\hat{z}$. The latest are constants vectors while $\hat{r}$ points towards the point that you're integrating (if I may say so). So it changes as you "sum" each $d\theta$ and $d\phi$ (if I may be so ruthless). So you can't pull it out of the integral as opposed to "inert" vectors like $\hat{x},\hat{y},\hat{z}$.

## 1. What is the potential of a spherical shell?

The potential of a spherical shell refers to the electric potential at a point outside or inside a spherical shell that is uniformly charged. It is calculated using the formula V = kQ/r, where k is the Coulomb constant, Q is the total charge of the spherical shell, and r is the distance from the center of the shell to the point of interest.

## 2. How does the potential of a spherical shell depend on distance?

The potential of a spherical shell is inversely proportional to the distance from the center of the shell. This means that as the distance increases, the potential decreases. This relationship follows the inverse square law, similar to the electric field of a point charge.

## 3. Can the potential of a spherical shell be negative?

Yes, the potential of a spherical shell can be negative if the total charge of the shell is negative. This means that the electric potential at a point inside the shell will be negative, while the potential at a point outside the shell will be positive.

## 4. What is the difference between potential inside and outside a spherical shell?

The potential inside a spherical shell is constant and does not depend on the distance from the center, as long as the point is within the shell. On the other hand, the potential outside the shell is inversely proportional to the distance from the center. Additionally, the potential outside is affected by the total charge of the shell, while the potential inside is not affected by the charge.

## 5. How is the potential of a spherical shell related to the electric field?

The potential of a spherical shell and the electric field are closely related. The electric field can be calculated by taking the negative derivative of the potential with respect to distance. In other words, the electric field is the rate of change of potential with distance. This means that where the potential is high, the electric field will be strong and vice versa.

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