The explacountry given to the answer of over question, was "Electric area is constantly perpendicular to equipotential surfaces".

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This idea was never before stated in the theory component of the book, so I wanted to know more about it.


If the field lines are not perpendicular to the surchallenge, then there is a component parallel to the surchallenge. If there is an electrical field component parallel to the surface, then a test charge placed on the surface would move alengthy the area line.

But the surface is an equipotential surchallenge, and also there is no pressure on a charge via no potential distinction. This offers a contradiction which means the initial presumption of field lines being not perpendicular to the surchallenge is wrong.

I like to make an analogy through gravity. If you think of equipotential lines as a topographical map. If you collection a ball on the hill, it will roll down the steepest slope at that point, which will certainly be perpendicular to the altitude lines.

The electric field is offered as the negative gradient of the potential. Lets take look at a conductor, which is characterized as having actually an equipotential surconfront. All charges reside on the surconfront of the conductor by interpretation, and given that that surface is equipotential, the charges distribute themselves such that the potential all over on the surconfront is equal. Therefore, the potential that arises because of a conductor is defined geometrically in regards to its surconfront. If you've had vector calculus, you may recontact that taking the gradient of a surface always gives you the vector normal to that surchallenge. Hence, considering that the electrical area is discovered by taking the gradient of the potential, the electric field is always perpendicular to the surchallenge on which the resource of the potential stays.

It simply... is. That's actually what provides it an equipotential surchallenge.

On Planet, if you desire to relocate in such a method that your gravitational potential energy does not change, you should relocate horizontally, perpendicular to the gravitational pressure. If you relocate up or down, you'll move along the gravitational area and also will gain or shed gravitational potential energy.

Electric potential works the same method.

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To develop off this mathematically, it means that the pressure area is conservative, so therefore it's the gradient of a scalar area and gradients of some scalar field are always perpendicular to the scalar field. In various other words the gradient is the normal to some scalar area.

It’s pure math: E = grad ϕ. Let γ be a curve in the equipotential surface ϕ = c then ϕ∘γ(t) = c therefore by the chain rule