# Electric Potential inside an insulating sphere

• baseballfan_ny
In summary, the potential difference between the sphere's surface and a point inside the sphere is -kfrac{Qr^2} {2R^3} + kfrac{3Q} {2R}.

#### baseballfan_ny

Homework Statement
A sphere of radius R carries a total positive charge Q distributed
uniformly throughout its volume. Find the electrostatic
potential inside the sphere.
Relevant Equations
Vf - Vi = - integral (E dot dl)
I used the potential at the surface of the sphere for my reference point for computing the potential at a point r < R in the sphere. The potential at the surface of the sphere is ## V(R) = k \frac {Q} {R} ##.

To find the potential inside the sphere, I used the Electric field inside of an insulating sphere from Gauss' Law, ## \vec E = k \frac {Qr} {R^3} \hat r ##.

The potential difference is given by ## \Delta V = -\int_{P_1}^{P_2} \vec E \cdot d \vec l ##

Using this I got $$V(r) - V(R) = -\int_{R}^{r} \vec E \cdot d \vec l$$

$$V(r) - V(R) = -\int_{R}^{r} k \frac {Qr} {R^3} \hat r \cdot d \vec l$$ where ## d \vec l ## points from R (outer radius) to r (inner) and is shown in the drawing below.

(I think this might be where I have a problem but I'm not sure why)

Because ##\hat r## and ##d \vec l ## point in opposite directions (they are anti parallel), I have the dot product $$\hat r \cdot d \vec l = | \hat r || d \vec l | \cos(\pi) = -dr$$ where dr is the magnitude of ## d \vec l ##.

This lead me to $$V(r) - V(R) = -\int_{R}^{r} k \frac {Qr} {R^3} -dr$$

$$V(r) - V(R) = \int_{R}^{r} k \frac {Qr} {R^3} dr$$

$$V(r) - V(R) = k \frac {Q} {R^3} \int_{R}^{r} r dr$$

$$V(r) - V(R) = k \frac {Q} {R^3} \frac {r^2} {2} - \frac {R^2} {2}$$

$$V(r) = k \frac {Q} {R^3} * (\frac {r^2} {2} - \frac {R^2} {2}) + V(R)$$

$$V(r) = k \frac {Q} {R^3} * (\frac {r^2} {2} - \frac {R^2} {2}) + V(R)$$

$$V(r) = k \frac {Qr^2} {2R^3} - k \frac {Q} {2R} + k \frac {Q} {R}$$

$$V(r) = k \frac {Qr^2} {2R^3} + k \frac {Q} {2R}$$

This is wrong; my answer implies that the potential would increase as you move away from the positively charged sphere. Also my book says the correct answer is ## V(r) = -k \frac {Qr^2} {2R^3} + k \frac {3Q} {2R} ##. I'm not exactly sure how/what went wrong, but I think it has to do with the dot product ## \vec E \cdot d \vec l ##. Any help would be much appreciated! Thanks in advance! Diagram showing the sphere of radius R with ## \vec E ## radially outwards. Also indicated ## d \vec l ## pointing from R to r and the positive ## \hat r ## direction pointing radially outwards.

It is easy to get confused about signs when integrating the 'wrong' way.
You know that V(r) should be greater than V(R), but on the RHS you appear to have three sources of negativity: ##\vec E.\vec{dl}##, the range from R to r and the actual minus sign.
Note that if we reverse the integration direction and correspondingly remove the minus prefix, it suddenly becomes positive. This is because the 'reversed' integration direction and inward direction of ##\vec {dl}## are really the same thing.

• baseballfan_ny
haruspex said:
It is easy to get confused about signs when integrating the 'wrong' way.
You know that V(r) should be greater than V(R), but on the RHS you appear to have three sources of negativity: ##\vec E.\vec{dl}##, the range from R to r and the actual minus sign.
Note that if we reverse the integration direction and correspondingly remove the minus prefix, it suddenly becomes positive. This is because the 'reversed' integration direction and inward direction of ##\vec {dl}## are really the same thing.

I think I understand. The LHS V(r) - V(R) is greater than 0, so we have to make the RHS consistent. Because r < R, switching integration order and removing the minus prefix would make RHS positive -- provided we also make ##\hat r## and ##d \vec l## in the same direction, such that ## \hat r \cdot d \vec l ## = dr (since I seem to be determining the direction of ## d \vec l ## by the integration bounds, and we just flipped those to [r, R] going radially outwards). So we get $$V(r) = k \frac {Q} {R^3} \int_{r}^{R} r dr + V(r)$$ $$V(r) = k \frac {Q} {R^3} ( \frac {R^2} {2} - \frac {r^2} {2}) + k \frac {Q} {r}$$ $$V(r) = -k \frac {Qr^2} {2R^3} + k \frac {3Q} {2R}$$, which is exactly the answer in my book.

I was thinking a bit about this problem during the day, and I came up with another interpretation for my problem that might help avoid all of the confusion I went through. I'm not sure if this is proper physics/math so correct me if I am wrong:

Looking at ## V(r) - V(R) = -\int_{R}^{r} k \frac {Qr} {R^3} \hat r \cdot d \vec l ## I have the positive direction of ## \hat r ## pointing radially outwards, so ##d \vec l##, which is the infinitesimally small element in the ## \hat r ## direction should also points radially outwards to be consistent with the coordinates. I can simplify the line integral ## V(r) - V(R) = -\int_{R}^{r} k \frac {Qr} {R^3} \hat r \cdot d \vec l ## as ## V(r) - V(R) = -\int_{R}^{r} k \frac {Qr} {R^3} dr ## since I made ## d \vec l## in the ##\hat r## direction. The negative line integral from R to r can be interpreted as walking from R to r while summing up all of the ## \vec E \cdot d\vec l## 's (which point in the same direction).

Basically I'm thinking that (to avoid confusion for myself) ## d \vec l ## should be determined by the choice of coordinate system, not the integration bounds. Is that an okay way to do it/a valid interpretation?

baseballfan_ny said:
## d \vec l ## should be determined by the choice of coordinate system, not the integration bounds.
That's not how I see it.
In a scalar integral, ##\int_a^bf(x).dx## means summing all bits like f(x).dx over intervals length dx from a to b. That is, if we were to sum just the dx we should get b-a.
If we switch the order to ##\int_b^af(x).dx## then we now need the sum of the dx to be a-b. So we have implicitly turned each dx into -dx, and that is why it flips the sign of the result.
The same applies to a vector integral. It is required that the sum of the element vectors equals upper bound minus lower bound. So reversing the bounds implies reversal of the elements; we don't need to reverse them explicitly as well:
##\int_{\vec a}^{\vec b}\vec F.\vec{dr}=-\int_{\vec b}^{\vec a}\vec F.\vec{dr}##

• baseballfan_ny
In the scalar product ##\vec E\cdot d\vec r##, ##|d\vec r|=-dr##, because you chose the integration interval to be from big to small.

• baseballfan_ny
archaic said:
In the scalar product ##\vec E\cdot d\vec r##, ##|d\vec r|=-dr##, because you chose the integration interval to be from big to small.
Yes, as I read post #1, @baseballfan_ny understood that.
The initial equation was ## V(r)−V(R)=−\int_R^r\vec E⋅\vec dl##.
Applying what you say above, ## \vec E⋅\vec dl## was evaluated as ## -|\vec E⋅\vec dl|## to give ## V(r)−V(R)=\int_R^r|\vec E⋅\vec dl|## and thence ## V(r)−V(R)=-\int_r^R|\vec E⋅\vec dl|##
But now we can see the sign is wrong.
My explanation is what you wrote but takes it a bit further: evaluating the integrand on the basis that the two vectors are in opposite directions is the same as reversing the range, not a separate operation. So we get
## V(r)−V(R)=+\int_r^R|\vec E⋅\vec dl|##

• archaic
I'm a bit confused what the fuss is about here... the electric field is defined piecewise:$$\vec{E}(r) = \begin{cases} \frac{kQr}{R^3} \hat{r} & r \leq R \\ \frac{kQ}{r^2} \hat{r} & r > R \end{cases}$$Use the definition of the potenial, ##\vec{E} = -\nabla \phi##,$$\phi(r') = -\int_{\infty}^{r'} \vec{E} \cdot d\vec{r} = - \left( \int_{\infty}^{R} \frac{kQ}{r^2} dr + \int_{R}^{r'} \frac{kQr}{R^3} \right) = \left[\frac{kQ}{r}\right]_{\infty}^{R} - \left[ \frac{kQr^2}{2R^3} \right]_{R}^{r'} = \frac{kQ}{2R} \left( 3 - \frac{r'^2}{R^2} \right)$$If you want to be pernickety you can specify that you're integrating over the trajectory ##\vec{r}(\lambda) = \lambda \hat{r}## from ##\lambda = \infty## to ##\lambda=r'##, then ##\frac{d\vec{r}}{d\lambda} d\lambda = \hat{r} d\lambda## points toward the centre of coordinates, but you don't need to worry about that.

If the signs still bother you, then as a nice heuristic, just think about the quantity ##\delta \phi = - \vec{E} \cdot \delta \vec{r}## for some small displacement ##\delta \vec{r}##.

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• baseballfan_ny
etotheipi said:
I'm a bit confused what the fuss is about here
It can be abstracted to this...
Consider ##\int_{\vec a}^{\vec b}\vec F.\vec{dx}##, where the natural direction of ##\vec {dx}## is that of ##\vec a-\vec b##, not of ##\vec b-\vec a## as implied by the limits.
Since the natural direction of ##\vec {dx}## meant that ##\vec F.\vec{dx}## would be negative, @baseballfan_ny replaced it with ##-|\vec F.\vec{dx}|##, then as a separate step reversed the limits and correspondingly flipped the sign again.
But the overall consequence of those steps is to end up with the wrong sign. So the puzzle was which step is wrong and why.

• baseballfan_ny
haruspex said:
Consider ##\int_{\vec a}^{\vec b}\vec F.\vec{dx}##, where the natural direction of ##\vec {dx}## is that of ##\vec a-\vec b##, not of ##\vec b-\vec a## as implied by the limits.

I think it's not a good idea to talk about a natural direction of ##d\vec{x}## without referring to an integration path. I suppose you're talking about a straight integration path between ##\vec{a}## and ##\vec{b}##.

If so, then the only possible right answer for the 'natural direction' of ##d\vec{x}## is parallel to ##\vec{b} - \vec{a}##, not the other way around! Any other way will give contradictory/wrong results.

haruspex said:
Since the natural direction of ##\vec {dx}## meant that ##\vec F.\vec{dx}## would be negative, @baseballfan_ny replaced it with ##-|\vec F.\vec{dx}|##, then as a separate step reversed the limits and correspondingly flipped the sign again.

If ##\vec{a}## is a point at infinite separation, and ##\vec{b}## a point inside the sphere, then the integral$$I = -\int_{\vec{a}}^{\vec{b}} \vec{E} \cdot d\vec{r} = -\int_{r_a}^{r_b} E_r dr$$Since ##d\vec{r}## is anti-parallel to ##\vec{E}##, I guess you could say ##\vec{E} \cdot \delta \vec{r} = - |\vec{E} \cdot \delta \vec{r}|##, or alternatively ##E_r \delta r = -|E_r \delta r|##. But to notice this does you no good other than to check that the integrand is indeed negative, and that the change in potential along the trajectory is positive.

It's a dead end, because I have no idea how you integrate something like ##\int |f(x) dx|## with the ##dx## inside the absolute value! That seems a bit gobbledygooky to me, so the whole question appears flawed • haruspex
etotheipi said:
I think it's not a good idea to talk about a natural direction of ##d\vec{x}## without referring to an integration path. I suppose you're talking about a straight integration path between ##\vec{a}## and ##\vec{b}##.

If so, then the only possible right answer for the 'natural direction' of ##d\vec{x}## is parallel to ##\vec{b} - \vec{a}##, not the other way around! Any other way will give contradictory/wrong results.
Yes, that is what I meant in post #4. The direction of the element vector is implied by the limits, not by some independent notion of the variable's inherent direction.

• etotheipi
Ah, okay. Forgive me for being unnecessarily flippant ... of course you had this under control from the start etotheipi said:
of course you had this under control from the start
I was having trouble wording it. Your posts certainly helped.

• etotheipi