Why electric field is always perpendicular to equipotential?

  • I
  • Thread starter David Day
  • Start date
  • #1
12
1
I have some understanding, but I'm not sure about how accurate it is:

Electrostatic force is given by F = qE, where F and E are both vector quantities. If the dot product of either side and the displacement vector Δs along an equipotential line is taken, the equation becomes

F⋅Δs = qE⋅Δs.

F and E are parallel; equipotential lines surround a charge radially. Since no work is done to move a charge along an equipotential line, F⋅Δs = 0. This means

qE⋅Δs = 0,

but q, E, and Δs are assumed to be nonzero. The only way to make the equation zero is to make E perpendicular to Δs (i.e., cosθ = 0) in all cases.

I would really appreciate a better explanation and pointing out of any errors in my understanding.
 

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,737
279
I have some understanding, but I'm not sure about how accurate it is:

Electrostatic force is given by F = qE, where F and E are both vector quantities. If the dot product of either side and the displacement vector Δs along an equipotential line is taken, the equation becomes

F⋅Δs = qE⋅Δs.

F and E are parallel; equipotential lines surround a charge radially. Since no work is done to move a charge along an equipotential line, F⋅Δs = 0. This means

qE⋅Δs = 0,

but q, E, and Δs are assumed to be nonzero. The only way to make the equation zero is to make E perpendicular to Δs (i.e., cosθ = 0) in all cases.

I would really appreciate a better explanation and pointing out of any errors in my understanding.
This is correct. Note that the case of a point charge is a special case and there is no need to talk about point charges in your derivation. What you showed is that in general, the E field will be perpendicular to equipotential surfaces (they really are surfaces). Of course, this is true for the equipotential surfaces around point charges but your proof is more general than that.
 
Last edited:
  • Like
Likes David Day

Related Threads on Why electric field is always perpendicular to equipotential?

Replies
1
Views
2K
  • Last Post
Replies
7
Views
1K
Replies
5
Views
841
Replies
2
Views
2K
Replies
4
Views
3K
Replies
4
Views
10K
Top