In a region containing an electric field, equipotential surfaces have the potentials of 𝑉 equals 100 volts for 𝑧 equals 0.00 meters, 𝑉 equals 200 volts for 𝑧 equals 0.50 meters, and 𝑉 equals 300 volts for 𝑧 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 𝐸, where 𝐸 is a vector having 𝑥-, 𝑦-, and 𝑧- components. We’re told in the problem statement various potentials corresponding to different values of 𝑧. Let’s start by sketching out the 𝑧-values and the corresponding potentials.
Here, is a three-dimensional plot with 𝑥-, 𝑦- and 𝑧-dimensions marked out. On the 𝑧-axis, we’re told potential values for three different values of 𝑧: 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 𝑧-direction, which we can call 𝐸 sub 𝑧, 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 𝐸 sub 𝑥 and 𝐸 sub 𝑦? All the data we have indicates that 𝑉, the potential, only changes in the 𝑧-direction. It doesn’t change at all, as far as we know, in the 𝑥 or 𝑦. So 𝐸 𝑥 and 𝐸 𝑦 are both zero based on the information given.
So what is 𝐸, the electric field that includes all 𝑥-, 𝑦-, and 𝑧- dimensions? The vector 𝐸 equals 𝐸 sub 𝑥, 𝐸 sub 𝑦, 𝐸 sub 𝑧. Since we’ve solved for each of these three components, we can now insert them to write out 𝐸.
The electric field in this region 𝐸 is zero 𝐸 sub 𝑥, zero 𝐸 sub 𝑦, 200 𝐸 sub 𝑧 volts per meter.