Concept Question - Electric Potential, and Electric Fields

[DMac]

I have a few questions about electric potential and electric fields.

If two points have the same electric potential, is it true that no work is required to move a test charge from one point to the other? Does that mean that no force is required as well?

Also, How much work is required to move a charged particle through an electric field if it moves along a path that is always perpendicular to an electric field line? How would the potential change along such a path?

Thanks in advance!

 

[EngageEngage]

The electric field is a conservative field so what does this tell you?

for part b, use the classical relations for work, with F = qE:

$$W = q\int_{C}\vec{E}\cdot d\vec{l}$$

so, what happens when $$d\vec{l}$$ and $$\vec{E}$$ are perpendicular?

 

[DMac]

I mean no disrespect, but I did not understand any of that because I'm still learning basic high school physics. (So, I don't know integration, or...lol anything, really.)

 

[DMac]

(All I got from the expression for Work is I know that if those two vectors are perpendicular, then the dot product is zero. But...yeah I don't understand what those two vectors are exactly. Sorry)

 

[EngageEngage]

E is the electric field vector, and dl is a small element of the path that you are taking. so when you integrate that you will get contributions of all the small work elements, dW to get your total work W. The path that you want to take is an equipotential path -- you should remember from your course that equipotential lines always lie perpendicular to the electric field, so the answer 0 is right. I hope this has cleared it up some, let me know if not

 

[DMac]

My textbook never covers equipotential lines =( ...One thing I don't understand is that I always thought that it would take no work to move a particle in the same direction as the field (isn't this true for magnetic fields?) ...so...lol I'm still confused.

 

[EngageEngage]

it depends on the field. In the case of electric fields, it does take work to move a particle along a field (whether it is you doing the work or the field, depends on the circumstances). with electric fields, if you move perpendicular to the field, however, it will require no work. In fact, this will always be the case with any conservative field, because the field is constructed from a potential, the electric potential (usually denoted V or phi). Have you taken anything with vector analysis, by chance? I might be able to explain this in a more mathematical way which could make this more clear.

 

[eok20]

It is true that if two points have the same potential, then no work is required to move something from one point to the other. This is analogous to gravity: picture, for example, a skater on a half pipe. If she's on one end it doesn't require any work to get to the other side (assuming no friction, air resistance, etc), she'll slide down half and that will give her enough energy to make it just to the top of the other part.

Work is defined, in general, to be the part of the force that is parallel to the motion of the object times the distance. Therefore if you're moving something perpendicular to a force, there is no component that is parallel so that the work is 0.

 

[DMac]

To EngageEngage:

I know very basic vector analysis, like dot product, cross product, projections, direction cosines, etc. But that's pretty much it.