Electric field is always perpendicular to the equipotential surface.

a. True
b. False

a: true.

Explanation:

We can define an equipotential surface as a surface where the potential at any point of the surface is constant.

For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.

Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.

Now we want to see if the electric field is always perpendicular to these equipotential surfaces.

You can see that in the two previous examples this is true, but let's see for a general case.

Now suppose that you have a given field, and you have a test charge in one equipotential surface.

So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.

But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.

Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.

The correct option is a.

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form