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At the midpoint of the line joining two equal point charges, the field is zero, and the electric field lines on this line joining the charges each begin at a charge and kind of end at the midpoint. Does that violate the rule that electric field lines (atleast those of the coulomb force) can only start/end at charges/infinity? If not, why not?

A follow-up question - In light of the above case, how can we prove that a point of zero electric field (a null point) in an "arbitrary electrostatic field configuration" is necessarily a point of unstable equilibrium?

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Does that violate the rule that electric field lines (atleast those of the coulomb force) can only start/end at charges/infinity?

No. Although the electric fields of each of the two equal charges vectorially equal zero at the mid point between the charges, the electric field lines do not "end". The lines are diverted from that space but still wind up terminating at a charge. The following link shows examples.

https://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Lines

The diagram below from the link demonstrates that the lines end at the charges.

A follow-up question - In light of the above case, how can we prove that a point of zero electric field (a null point) in an "arbitrary electrostatic field configuration" is necessarily a point of unstable equilibrium?

Not quite sure what you mean by "prove", but clearly the field strength, which is proportional to the density of the field lines, is zero only at the mid point. A charge at any position not exactly at the mid point will experience a force and accelerate. Perhaps in that sense the situation of a charge at the mid point can be considered unstable.

Hope this helps.

enter image description here

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  • $\begingroup$ What about the particular field line on the line joining the charges? That doesn't get diverted in any direction. $\endgroup$ – Green05 Sep 21 '19 at 18:04
  • $\begingroup$ @Green05 the vector sum of the field of each charge equals zero there. So there are no field lines that, as you say, “kind of end at the midpoint”. The diagram doesn’t show a line from each charge that ends at the middle. I thought that was your point. $\endgroup$ – Bob D Sep 21 '19 at 20:04
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Does that violate the rule that electric field lines (at least those of the coulomb force) can only start/end at charges/infinity?
In the example you have given the answer is -no.

Your diagram shows the result of the addition of the electric fields due to two negative charges and it so happens that at one point the addition of the electric fields produces a net electric field is zero.
In between the two charges and along the line joining the two charges there are two electric fields, one from each charge and at one point on that line the net electric field is zero.

In light of the above case, how can we prove that a point of zero electric field (a null point) in an "arbitrary electrostatic field configuration" is necessarily a point of unstable equilibrium?

Consider the electric potential due to one negative charge, located at $(0,0)$, which is shown diagrammatically on the left hand side.
When another potential due to a negative charge, located at $(1,0)$, is added to the potential due to the charge at $(0,0)$ the potential diagram is as shown on the right hand side.

enter image description here

Consider a positive charge located at position $(0,0)$.
A small displacement along the x-axis will reduce its potential energy and it will feel no inclination to go back to position $(0.5,0)$ - unstable equilibrium.
However a small displacement along the y-axis will increase its potential energy and the charge will move back to position $(0.5,0)$ - stable equilibrium.
The position $(0.5,0)$ is called a saddle point.
There is no way of positioning the two charges so a not to hav a saddle point between them.

The reasoning can be done in terms of forces (equal to minus the potential gradient.
Moving away from position $(0.5,0)$ along the x-axis produces a net force on a positive charge away from the origin (unstable equilibrium) and movement along the y-axis will result in the positive charge have a force on it towards position $(0.5,0)$.

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