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In Concepts of Physics by Dr. H.C.Verma, in the chapter on "Capacitors", in page 146, under the topic "Calculation of Capacitance" for a "Spherical Capacitor" the following is given which is a part of derivation where the potential difference between the positively charged inner shell $B$ (radius $R_1$) and negatively charged outer shell $A$ (radius $R_2$) is calculated:

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$$V=V_+-V_-=-\int_A^B \vec E.d\vec r \tag1$$

$$=-\int_{R_2}^{R_1}\frac{Q}{4\pi\epsilon_0r^2}dr \tag2$$

$$=\frac{Q}{4\pi\epsilon_0}\left(\frac 1 {R_1}-\frac 1 {R_2}\right)=\frac{Q(R_2-R_1)}{4\pi\epsilon_0R_1R_2}.\tag3$$

Equation $(1)$ is the mathematical form of the definition of potential difference between two points. But in step $(2)$, the dot product is solved (or in other words vector notation is converted to scalar notation) and this is where my doubt starts.

On moving from the outer shell $A$ to the inner shell $B$, the vector $d\vec r$ points towards the centre whereas the electric field is radially outward. This implies $\vec E$ and $d\vec r$ are $\pi$ radian apart (antiparallel) and so the dot product is negative. But the author doesn't seem to include this negative sign in step $(2)$. In order to check whether the final result given by $(3)$ is correct or not, I determined the potential difference between the shells by an alternate method*. I found it to be same as that of $(3)$. But here, if I considered the negative sign in $(2)$ then I'll get the same magnitude of the potential difference but it'll be of opposite sign to that of $(3)$. So, how did the author obtain the correct result even after neglecting the negative sign due to the dot product of the antiparallel vectors?


*The alternate method I used to determine the potential difference is outlined by this answer which I got when I searched to get my doubt cleared. Please note that the question/answer Problem in Proving the potential due to concentric shells between the shells didn't clarify my doubt on the sign inconsistency discussed in my question.

Image constructed by me with the help of diagram 31.5 from the book mentioned.

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This is a confusion I had when I did these problems. We are integrating from the outer shell of radius $R_2$ to the inner shell of radius $R_1 < R_2$. You're right then along such an integration path, the $d \vec{r}$ points "inwards", while $\vec{E}$ points "outwards", so their dot product is negative. But, this negative sign is already being taken into account when we write \begin{align} \int_{R_2}^{R_1} \dfrac{Q}{4 \pi \varepsilon_0 r^2} \, dr \tag{$*$} \end{align} Notice that the integrand $\dfrac{Q}{4 \pi \varepsilon_0r^2}$ is positive (since $Q>0$), BUT, we are integrating from $R_2$ to $R_1$ (from higher limit to lower limit).

In other words, in $(*)$, we are integrating a positive function, but "in the opposite direction", hence, the integral in $(*)$ is actually negative, which is exactly what we expected: $\int \vec{E} \cdot d \vec{r}< 0$ for the chosen path, based on the physical argument that the vectors point in opposite directions.


So, what is going on is that the fact that $d \vec{r}$ points "inwards" is already taken into account by the limits of integration. (for example, compare $\int_1^2 x^2 \, dx$ vs $\int_2^1 x^2 \, dx$)

While this is a good question to think about, I have the following suggestion for the future: if for example you work in spherical coordinates, always write \begin{align} \vec{E} = E_r \hat{r} + E_{\theta} \hat{\theta} + E_{\phi} \hat{\phi} \quad \text{and} \quad d\vec{r} = dr\, \hat{r} + r \, d \theta\, \hat{\theta} + r \sin \theta \, d \phi\, \hat{\phi} \end{align} so that the dot product is \begin{align} \vec{E} \cdot d \vec{r} = E_r\, dr + E_{\theta} r \, d \theta + E_{\phi} r \sin \theta \, d \phi \end{align} and carefully set up the limits of integration, and let the limits of integration take care of the sign of everything, because it is very dangerous to use these physical arguments to get the signs correct in integration, because we might unknowingly introduce extra minus signs.

But of course, at the end of a calculation, always check with basic physical intutition.


If this is still unconvincing, consider the following very simple example of line integration: let $\vec{F}(x,y,z) = \hat{x}$, be a constant unit vector field pointing in positive $x$ direction. Consider the straight-line path $C$ which starts at the point $(2,0,0)$ and ends at $(1,0,0)$, and now I ask you to compute $\int_C \vec{F} \cdot d \vec{r}$.

Here, $d\vec{r}$ points in $-\hat{x}$ direction, while $\vec{F}$ points in $\hat{x}$ direction. So, of course, we expect the integral to be negative. But, if we naievely put $\vec{F} \cdot d \vec{r} = \hat{x} \cdot (-dx \, \hat{x}) = -dx$, and if you also put the limits of integration as $\int_2^1$, "because $C$ starts at $x=2$ and ends at $x=1$", then you'll incorrectly compute \begin{align} \int_C \vec{F} \cdot d \vec{r} &= \int_{2}^1 (-dx) = \int_1^2 \, dx = 2-1 = 1 > 0 \end{align} So, we found the integral is positive when it should be negative. So, clearly, we have added an extra minus sign.

The thing to note is if you write $d\vec{r} = - dx \, \hat{x}$, then you need to reverse the integration limits as $\int_1^2$, even though $C$ starts from $x=2$ to $x=1$. To me this is very confusing, so I rather write \begin{align} \int_C \vec{F} \cdot d \vec{r} &= \int_{x=2}^{x=1} \hat{x} \cdot (dx \, \hat{x} + dy \, \hat{y} + dz \, \hat{z}) \\ &= \int_{x=2}^{x=1} dx \\ &= 1 - 2 \\ &= -1 \end{align} This way, we are less likely to make sign errors.

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  • $\begingroup$ +1: Thank you for your answer. As you pointed out in $(2)$ there should not be any vector. It was a mistake on my part which went unnoticed. I'm sorry if that confused you. I hope the statement in parenthesis below $(*)$ regarding this could be removed as I fixed the issue in the question. $\endgroup$ – Guru Vishnu Dec 17 '19 at 10:01
  • $\begingroup$ @M.Guru Vishnu, you're welcome, and no it didn't cause any confusion, I just wanted to let you know (I didn't know if it was just a typo or an actual conceptual issue hence I just wrote it down) $\endgroup$ – peek-a-boo Dec 17 '19 at 10:05

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