Why is the electric potential at a distance of $R$ from a point charge $q$ equal to $\frac{-q}{4\pi\varepsilon_0 R}$

Question

My textbook mentions that the electric potential at distance of $R$ from a point charge $q$ will be given by $\dfrac{-q}{4\pi\varepsilon_0 R}$. I don't understand why the negative symbol appears here.

As I understand it, the electric potential there should be the amount of work that needs to be done by an external agent to bring a unit charge from infinity to that point without accelerating it (i.e. without a change in kinetic energy). How I learnt to derive it is given below :

Let $\vec{F_2}$ be the force applied by the external agent at a point between $R$ and $\infty$. It is opposite in direction to $\vec{F_e}$, which is the electrostatic force at that point. Their magnitudes are almost equal. $|\vec{F_2}| = |\vec{F_e}| + dF$. Here, $dF$ is the negligible 'extra' force. It accelerates the test charge negligibly and hence, there is a negligible change in kinetic energy which can be ignored. Now, the total work done by the external agent to bring the test charge ($q_2$, let's say), from infinity to a distance of $R$ from $q$ will be : $$\int_\infty^R \vec{F_2}.\vec{dr}$$ Now, as we have established that magnitudes of $F_e$ and $F_2$ are approximately equal, we can write this integral as : $$\int_R^\infty \vec{F_e}.\vec{dr} = \dfrac{qq_2}{4\pi\varepsilon_0}\int_R^\infty \dfrac{dr}{r^2} = \dfrac{qq_2}{4\pi\varepsilon_0 R}$$ Now, this is the electric potential energy $(U)$ possesed by $q_2$ placed at a distance of $E$ from $q$. Now, $V = \dfrac{U}{q_2} = \dfrac{q}{4\pi\varepsilon_0 R}$

I don't see how the negative symbol can appear here, especially because the work done by the external agent would be positive since the displacement occurs in the direction of the force that it applies. Also, both the charges are given to be positive.

So, is my definition of electric potential wrong or is it something else?
Thanks!

Answer 1

The electrical potential due to a charge $q$ a distance $R$ away is $$V = \frac{1}{4\pi\epsilon_0}\frac{q}{R}.$$ There is no minus sign, you should double check how your book defines the electric potential. It might just be a mistake, in which case I would be very careful about consulting that book about anything.

Answer 2

Potential is defined as $$V(r) = -\int_\vec{O}^\vec{r}\vec{E}\cdot d\vec{l} \\ \int_\vec{a}^\vec{b}(\vec{\nabla}V)\cdot d\vec{l} = -\int_\vec{a}^\vec{b}\vec{E}\cdot d\vec{l}\\ \vec{E} = -\vec{\nabla}V$$

The gradient points in the direction of the biggest change. A positive charge has a gradient towards the origin of the coordinate system

Now, this is for gravitational potential but the same holds for electric (couldn't find the electric one, just image it says electric potential and the $y$ ordinate is $V$ and $x$ abscissa is $displacement$. Potential above $0$ is for positive charges and the gradient points towards the origin while below $0$ (for negative charge), the gradient points outwards out of the origin. This is ofc along the lines of the potential.

!! We put the minus sign so that the field comes out of the positive charge !! (and negative to be the drain)