why are electric fields perpendicular with the equipotential lines?

Question

its all about electrostatics in physics.

Answer 1

The electric fied represents lines of electric force that are exerten on a point charge in a particular point in space. The force is the same if that point is located equal distance from the charge that causes the electric field in the first place.

E=kq/r^2 (If r- does not change then neither is E)
E -field intensity
k - proportionality constant( you know exactly what it is)
q - charge taht has crested that field
r - the distance to that point in space we are talking about.

Finnaly anywhere around this charge the force will be the same or electric fiel will have the same value and the same electric potential creating an equipotential ring or actualy a sphere. The force is directed radialy outward and is perpendicular to the circumferennce of the equipotential rings or actualy ( equipotential sphere).

Does that makes scense?
See references as well and have fun!

Answer 2

Imagine they were not.
Then there would be a component of the electric field parallel with the equipotential line.
So moving a charge on the equipotential line against that force would require you to do work.
Hence there is a difference in potential along the equipotential line. This is offcourse a contradiction.

So the component of the electric field along the equipotential lines should be zero, which means that the electric field has to be perpedicular to those lines.

Answer 3

Take 2 adjascent factors A and B on the equipotential line. A fee placed at A won't circulate to B because of the fact the potentials are an identical. it is the sector in this course is 0. yet A is having some ability and there might desire to be a field in some course such that its factor alongside AB is 0. the only course is perpendicular to AB.