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According to the formula, $$V = - \int \vec E \cdot d\vec r.$$

The electric potential of a point charge is $\frac{KQ}R$? However, when I perform the integration, my answer is $-\frac{KQ}R$.

Since the direction of the electric field is opposite of a test charge that moves from infinity to $R$, the electric field dot product $d\vec r$ is a negative. Furthermore, the integral of $E$ is also a negative, so totally there are three negatives, and the final term is $-\frac{KQ}R$.

Could anyone clarify any mistakes I made?

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  • $\begingroup$ Note that in my answer below if you want to make $d\vec r$ negative you also have to change the sign of the force that you are exerting on the unit positive charge and you still get a positive answer as two negatives when multiplied together give a positive. $\endgroup$ – Farcher Feb 7 '16 at 11:47
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I am not sure but I think: $V=\int_r^{\infty}\vec{E}d\vec{r}$, that is why it is positive.

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-ve Sign which comes during integration is due to opposite directional motion of test charge.

You can't consider it twice, once during integration and next during antiparallel motion.

Hint: Why do you think - ve sign comes during integration? Since you know Electric field was positive, How integrating it could be -ve.

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The connection between electric potential and E-field strength can be conceptually difficult usually because some important ideas are forgotten.

The first thing is the definition of potential.
Potential is defined as the work done by an external force in taking unit positive charge from a arbitrarily chosen zero of potential to the point.

For convenience I will ask you, the reader, to apply the external force.

enter image description here

So you start at point $A$, the arbitrary zero of potential and take a unit positive charge $+!$ on a walk to point $B$.

At some time in your walk you reach a point $X$ where the force on the charge is $\vec E$. To keep the charge moving along the path you wish it to take you are applying and equal and opposite force $-\vec E$.
To take the next step of length $d\vec r$ forward you have to do an amount of work $(-\vec E) \cdot d \vec r$.

So adding all the steps together in going from $A$ to $B$ the amount of work that you have to do is $\int_A^B (-\vec E) \cdot d \vec r$.

So now look at your point charge derivation.

The convention is that the zero of potential is at infinity. So this is point $A$.
When you arrive at point $X$ with your unit positive charge you feel a force on the charge $\frac {kQ}{r^2}$ which is in a direction away from the charge. You must apply an equal and opposite force on the charge in the direction that you are moving (towards the charge $Q$) to make your next step. So you do positive work and the potential of the unit positive charge increases as you make that step.

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  • $\begingroup$ But if electric potential = -Integral E dot dr I don't see how that can be an external force, since the force or field is the system itself. Also, change in potential energy is the negative work done by a conservative force. An external force is not conservative. $\endgroup$ – Goldname Feb 7 '16 at 15:14
  • $\begingroup$ The system is the unit positive charge and whatever is producing the E-field. You provide the external force which at all times is equal in magnitude and opposite in direction to the force produced on the unit positive charge by the E-field. That work done by you is independent of the path you take. The important thing is to look at the definition of potential. Another way which potential is defined is in times as minus the work done by the E-field. $\endgroup$ – Farcher Feb 7 '16 at 15:38

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