### Video Transcript

In a region containing an electric field, equipotential surfaces have the potentials of π equals 100 volts for π§ equals 0.00 meters, π equals 200 volts for π§ equals 0.50 meters, and π equals 300 volts for π§ equals 1.00 meters. What is the electric field in this region?

We can call the electric field that we want to solve for in this region πΈ, where πΈ is a vector having π₯-, π¦-, and π§- components. Weβre told in the problem statement various potentials corresponding to different values of π§. Letβs start by sketching out the π§-values and the corresponding potentials.

Here, is a three-dimensional plot with π₯-, π¦- and π§-dimensions marked out. On the π§-axis, weβre told potential values for three different values of π§: at the origin, at 0.50 meters and at 1.00 meters. At those points, we have equipotential surfaces of 100, 200, and 300 volts, respectively. If we consider this progression as a gradient caused by an electric field in the π§-direction, which we can call πΈ sub π§, we can see from these points that over a distance of one meter, the electric potential increases by 200 volts. So the electric field along this dimension is 200 volts per meter.

Now what about πΈ sub π₯ and πΈ sub π¦? All the data we have indicates that π, the potential, only changes in the π§-direction. It doesnβt change at all, as far as we know, in the π₯ or π¦. So πΈ π₯ and πΈ π¦ are both zero based on the information given.

So what is πΈ, the electric field that includes all π₯-, π¦-, and π§- dimensions? The vector πΈ equals πΈ sub π₯, πΈ sub π¦, πΈ sub π§. Since weβve solved for each of these three components, we can now insert them to write out πΈ.

The electric field in this region πΈ is zero πΈ sub π₯, zero πΈ sub π¦, 200 πΈ sub π§ volts per meter.