In a particular region, the electric potential is given by 𝑉 equals negative 𝑥 𝑦 cubed 𝑧 plus five times 𝑥 times 𝑧, where 𝑥, 𝑦, and 𝑧 correspond to distances in meters. Find the electric field in volts per meter at a point in this region that has the coordinates 5.00, 3.00, 1.00.
Here we’re given an equation for electric potential in terms of distances in meters in three dimensions, 𝑥, 𝑦, and 𝑧. Electric potential is a scalar value. That is, it has no direction associated with it. But electric field, which is in units of volts per meter, is a vector value. And it’s that we want to solve for at a particular coordinate point using this equation for electric potential.
That then raises the question, what’s the connection between electric field and electric potential? We can start to understand how these two quantities are connected by considering their units. We’re told that electric field, 𝐸, in this case is in the units of volts per meter, while electric potential we know is given in units of volts. This suggests that if we take our electric potential in units of volts and take the derivative of that value with respect to distance, then we’ll end up with electric field, 𝐸.
Now that’s just a suggestion based on the similarity in units. We haven’t proven anything yet, but it turns out that our assumption is correct. When our potential is a function of only one variable, say 𝑉 as a function of 𝑥, then the corresponding electric field is equal to the derivative of 𝑉 with respect to 𝑥 in the 𝑥-direction, and it’s the negative of all that.
You might wonder why the negative sign. The reason has to do with where we conventionally set potential equal to zero at ∞. This means that if we have a positive charge and we set our potential at ∞ to be equal to zero, then as potential moves in from ∞ towards our charge, it will be increasing. But as we consider the electric field created by this positive charge, we know that as we get farther and farther away from the charge, it will decrease. It will move in the opposite direction of potential. And that’s why we have a negative sign in this expression.
But back to the question at hand, what if, instead of having a one-dimensional potential, we have a three-dimensional potential, one that’s a function of 𝑥, 𝑦, and 𝑧? In that case, to solve for electric field based on this potential, we’ll want to take not one but three derivatives, and they’ll be partial derivatives.
Here’s what it looks like mathematically. With a three-dimensional electric potential, the corresponding electric field is equal to negative the partial derivative of 𝑉 with respect to 𝑥 in the 𝑥-direction plus the partial derivative of 𝑉 with respect to 𝑦 in the 𝑦-direction plus the partial derivative of 𝑉 with respect to 𝑧 in the 𝑧-direction. It’s this formulation that we’ll use to solve for electric field based on a potential that depends on three different variables.
So let’s start writing all this out. Our electric potential, which is a function of 𝑥, 𝑦, and 𝑧, is equal to negative 𝑥 𝑦 cubed times 𝑧 plus five 𝑥𝑧, where 𝑥, 𝑦, and 𝑧 correspond to distances in meters. We want to calculate electric field based on this. And to do that, let’s go through one by one and take these three partial derivatives of our electric potential. Once we’ve done that, we can piece all three parts together to solve ultimately for electric field, 𝐸.
We’ll start out then with the partial derivative of 𝑉 with respect to 𝑥, that is, differentiating our electric potential with respect to the variable 𝑥. Remembering our differentiation rules, we see that works out to negative 𝑦 cubed times 𝑧 plus five times 𝑧. That’s the first of three derivatives we’ll take.
Next, we’ll do d𝑉 d𝑦. Differentiating our potential with respect to 𝑦, we see we’ll make use of the chain rule in order to differentiate this first term. The derivative of this first term is negative three times 𝑥 𝑦 squared times 𝑧, and the derivative of the second term with respect to 𝑦 is zero. That’s our second derivative.
And now we’ll move on to the third and final one. The derivative of 𝑉 with respect to 𝑧 is negative 𝑥 𝑦 cubed plus five times 𝑥.
We’ve now got all three of our terms which form the components of the electric field vector. Before we combine them according to this relationship, let’s plug in the given values at which we want to solve for the electric field. Here we have 𝑥 corresponding to 5.00, 𝑦 corresponding to 3.00, and 𝑧 corresponding to 1.00. Let’s apply these three values to our first partial derivative term, d𝑉 d𝑥.
Plugging in for the given 𝑦 and 𝑧 values, when we calculate this expression, it comes out to negative 22.0. That’s our 𝑥-component of the electric field. Now let’s consider the 𝑦-component, d𝑉 d𝑦. Plugging in for 𝑥, 𝑦, and 𝑧 here, 3.00 squared is 9.00 multiplied by 5.00 is 45.0. And all that multiplied by three times 1.00 is equal to negative 135. That’s the 𝑦-component of the electric field.
And finally, on to the 𝑧-component, plugging in for those values, 3.00 cubed is 27.0 multiplied by 5.00 is 135 with a minus sign in front. And negative 135 plus five times 5.00 all comes out to negative 110.
We’ve made great progress. We now have the 𝑥-, 𝑦-, and 𝑧-components of our electric field, 𝐸. All we need to do now is arrange these coordinates in their proper relationship, and we won’t wanna forget the minus sign leading off all these values. Here then is how we can write out our electric field, 𝐸.
We can say it’s equal to positive 22.0 volts per meter in the 𝑥-direction plus 135 volts per meter in the 𝑦 plus 110 volts per meter in the 𝑧-direction. Or another way to write this is to use parentheses and write 22.0𝐢, 135𝐣, 110𝐤 volts per meter. That’s the electric field based on this electric potential at the given coordinate location.