Why is the electric potential from a positive sphere not negative? [duplicate]

Question

Say we have a positive point charge located somewhere. Then the electric field due to this point charge is $\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r} $ , now considering potential is $V = -\int_{a}^{b}\mathbf{E}\cdot d\mathbf{s}$ then the potential at an arbitrary point $r_1$ away from the charge(considering $V_\infty=0$) is $$V = -\int_{\infty}^{r_1}\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r} \cdot d\mathbf{s}$$. Now considering we are going from infinity to $r_1$, $d\mathbf{s}$ should be pointed radially inward, while the electric field should be pointed radially outward giving a $cos(180)$ so I get

$$V = -\int_{\infty}^{r_1}\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}cos(180) dr$$

$$= \int_{\infty}^{r_1}\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2} dr$$ which ends up being $$V=-\dfrac{Q}{4\pi \epsilon_0r_1}$$.

What am I missing here?

Answer 1

You have a really wonderful doubt. You did the calculations right, except for the dot product. It is a error easy to overlook.(Note: bold means vectors). Dot product E.dr is equal to Edrcos$\theta$. But, you have to note here that E and dr are magnitudes and must be positive. Since you come from infinity, you get that dr is negative. So, its magnitude is -dr.

Therefore E.dr = |E||dr|cos$\theta$ = E(-dr)cos$\theta$.

This is the minus sign you missed in your calculations.

You can also think physically that to bring 1C charge from infinity to some distance, you have to do some work, therefore the potential increases.

Also in some cases if given, the potential of infinity may not be 0, then you have to use change in potential to find potential at some point.

Answer 2

The potential energy is the work done on the unit charge as it is moved from $\infty$ to $r$. That is, suppose you are pushing the charge inwards, then the change in PE is the work you do. That means the force we need to calculate the work is directed in the $-r$ direction because it is the inwards force that you are exerting on the charge.

The rationale for this is that if the PE increases that means the energy of the system has increased, and that means energy must have been added to the system from an external source. In this case that external source is you, and the energy added to the system is the work that you are doing on it.

What you have calculated is the work done by the electric field, not the work done by an external agent on the field. The work done by the field is indeed negative, which means the energy of the electric field is increasing.

Understanding the sign of work is notoriously confusing even for those of us who have been physicists all our life. The approach I take is to look at how the energy is moving. If the PE of a system is increasing then energy must be added to the system from an external source. Conversely if the PE is decreasing then the system must be transferring energy from the system to some external sink.