When I calculate the work required to bring a charge from an infinite distance away to the midpoint of a dipole, I obtain zero. I know the result is correct because of the calculations. What I don't understand is the physical meaning of this. I've been told that it would be helpful to think about it considering the direction of the electric field, but I find this confusing, since I can't see the connection between both things.
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
What I don't understand is the physical meaning of this.
The physical meaning comes directly from the definition of work. Work is the dot product of force and displacement, or
$$dW=\vec F\cdot d\vec s = Fds\cos\theta$$.
In the case of an electric charge and electric field
$$dW=Q\vec E \cdot d\vec s=QEds\cos\theta$$
Where $\theta$ is the angle between the electric field $\vec E$ and the displacement $d\vec s$.
When a charge is brought from infinity (point A in the figure below) along a path precisely between the charges of a dipole, the displacement of the charge is perpendicular to direction of the field lines of the dipole. Since the electric field of the dipole is perpendicular to the path, the force exerted by the field on the charge is perpendicular to its displacement. Thus the work is zero.
Hope this helps.
-
$\begingroup$ It's very much like a ball moving along the floor in a physics class that ignores friction. If it never goes up work is never done. Why movement is happening at all doesn't need to be explained. Why shouldn't movement be happening? $\endgroup$ Commented Sep 5 at 13:30
-
$\begingroup$ @candied_orange “why movement is happening at all doesn’t need to be explained” But, it is explained by Newton’s first law. “Why shouldn’t movement be happening”. What movement are you referring to? $\endgroup$– Bob DCommented Sep 5 at 13:41
-
$\begingroup$ "... bring a charge from ..." <- That movement. The movement that this trick question implies requires work to be done. That movement does not require work. Or explanation. Why am I explaining this? $\endgroup$ Commented Sep 5 at 16:29
$\begingroup$
$\endgroup$
If you move the charge perpendicular to the electric field lines, no work is done. This is because work $W$ is defined as $$ W = \int_{\vec{r}_0}^{\vec{r}_1} \vec{F} \cdot d\vec{s} = q\int_{\vec{r}_0}^{\vec{r}_1} \vec{E}\cdot d\vec{s}, $$ where you move the charge $q$ from $\vec{r}_0$ to $\vec{r}_1$. Note the dot product between the electric field $\vec{E}$ and the infinitesimal line element $d\vec{s}$ that is zero iff they are perpendicular to each other. This means if you move the charge always perpendicular to electric field, namely on an equipotential line, no work will be done. If you draw the field lines, you can find an equipotential line for every charge position to midpoint of the dipole.
Another way of thinking about this is that any work you do against one dipole charge is later gained by repulsion of the same dipole charge and/or attraction of the opposite one (depends on the chosen path).
Note that all paths with the same start and end point require the same amount of work since the force field is conservative.
answered Sep 4 at 19:03
$\begingroup$
$\endgroup$
1
Work done to bring a charge from infinite distance to mid-point is zero as the electric potential is zero. The electric potential at mid point is zero as both the charges (negative and positive) are at midpoint of dipole and cancels each other, therefore no work is needed to move the charge in potential energy.
-
3$\begingroup$ The work done is zero because the change in potential is zero. The potential is zero at infinity and is also zero at the midpoint. The absolute value of the potential being zero is meaningless. Only the change matters. $\endgroup$– hftCommented Sep 4 at 19:06