### Video Transcript

Three 5.0-microcoulomb charges are fixed at the vertices of a triangle that has 3.0-centimeter long sides. Find the total electric potential energy of the system.

First thing to notice here is we’re asked to solve for the total electric potential energy of the system; that’s important to differentiate from electric potential. The second thing is that because these three charges are at the vertices of a triangle that has equal side lengths, that tells us that each one of our three charges, which we’re calling 𝑄 since they’re all the same, are at distance of 𝑑 from any of the other two charges.

Now that we have a sense for the geometry of this setup, let’s consider the electric potential energy of this system of three charges. In general, if we have two charges 𝑄 one and 𝑄 two, then the electrical potential energy of that system of charges — the two point charges themselves — is equal to the product of their charges times Coulomb’s constant all divided by the distance between them. And again, this relationship is for two point charges.

When we look over at our triangle of charges though, we see that we have three point charges rather than two. But thinking it over, we could think of these three charges as three pairs of electrical charges.

What makes that convenient is that all of our charges have the same sign and same magnitude and are separated by the same distance 𝑑. If we were to write out the electric potential energy of just one pair of these three charges, we’ll call it 𝑢 sub 𝑝, then based on our guiding equation, we will write it this way.

We would say that 𝑢 sub 𝑝 is equal to 𝑘 times 𝑄 squared divided by 𝑑. 𝑄 is being squared because again each of our charges has the same sign and magnitude. So far so good, but remember we don’t wanna solve for the electric potential energy of one pair of these three charges, but for the entire system.

But here’s the really great thing: because all three of our pairs of charges have the same charge sign and value and are separated by the same distance, we can say that the total electric potential energy of this system — we’ll call it 𝑢 sub 𝑇 — is equal to three times the electric potential energy of one pair of charges in the system.

To solve for that, we just need to plug in for the values of 𝑄, 𝑑, and Coulomb’s constant 𝑘. We can write that constant as 8.99 times 10 to the ninth newton meter squared per coulomb squared. And beyond that, we can write capital 𝑄, our given charge value, as 5.0 times 10 to the negative sixth coulombs. And the distance 𝑑 which is given as 3.0 centimeters we know to be equal to 0.03 meters, the value we’ll use in the calculation.

When we plug all these values into our expression for the total electric potential energy, notice what happens to the units here. We see that the units of coulomb squared cancel out and one unit of meters cancels out from this expression too. We’re left with the units of newtons times meters, which is the unit of energy joules. This is a good sign because we’re solving for an energy, a potential energy.

To two significant figures, all this simplifies to 22 joules of electrical potential energy. That’s the total electric potential energy of this system of three charges.