1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It says that the negative of the work done by internal force in bringing a unit positive charge from infinity to that point is the electric potential at that point. I don't quite get what that means though. But, I saw a lecture of Professor Walter Lewin and used another book, and just as you say, the negative symbol was not there. It has the negative sign in the formula for electric potential energy too. It just mentions that the electric potential energy will be negative of work done in bringing test charge from infinity to that point keeping the other charge fixed but doesn't tell why... $\endgroup$ – Rajdeep Sindhu Jul 3 at 9:00
  • $\begingroup$ That definition is correct if "internal force" means the force applied on the unit charge by the electric field (i.e. the existing charges). For a single positive charge, this work would be negative, so the potential would be positive. If there is no mistake in the book, you might be quoting the equation out of context. $\endgroup$ – Puk Jul 3 at 9:06
  • $\begingroup$ But the formula shouldn't be different, right? $\endgroup$ – Rajdeep Sindhu Jul 3 at 9:07
  • $\begingroup$ As I said, there should be no minus sign in the equation for the potential. $\endgroup$ – Puk Jul 3 at 9:08
-1
$\begingroup$

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
Potential

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)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What does $\nabla$ mean here? This is the first time I've seen it. $\endgroup$ – Rajdeep Sindhu Jul 3 at 9:11
  • 1
    $\begingroup$ This is just wrong. Not only does the equation make no sense, the figure refers to gravitational potential, which is negative for a point mass, so it is confusing. $\endgroup$ – Puk Jul 3 at 9:11
  • $\begingroup$ Your definition is wrong. It should be $-\int_\vec{a}^\vec{b}\vec{E}\cdot\,d\vec{l} = V(\vec{b})-V(\vec{a})$ or equivalently $\vec{E}=-\vec{\nabla}V$. $\endgroup$ – Thomas Fritsch Jul 3 at 10:02
  • $\begingroup$ True, I made a mistake, forgot one more line in the equation. Furthermore, I have made a remark that this picture is correct if you remove the equations from them and replace grav with electric. $\endgroup$ – Dominik Car Jul 3 at 22:47
  • $\begingroup$ @RajdeepSindhu en.wikipedia.org/wiki/Del it's just a short way of writing derivation. Nabla symbol takes potential $V$, derivates it by $x$,$y$,$z$, multiplied by $\hat{x}$,$\hat{y}$,$\hat{z}$ respectively. (potential is a scalar, nabla is a vector, so the electric field is a vector). khanacademy.org/math/multivariable-calculus/… $\endgroup$ – Dominik Car Jul 4 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.