Why electric field is always perpendicular to equipotential?

[David Day]

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.

 

[nrqed]

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.