Video: Determining the Force on a Charge due to a Second Charge

Point charges 𝑄₁ = 0.0050 nC and 𝑄₂ = 0.020 nC are fixed at π‘Ÿβ‚ = (9.00𝑖 + 3.50𝑗) m and π‘Ÿβ‚‚ = (6.50𝑖 βˆ’ 2.00𝑗) m respectively. What is the magnitude of the electric force on 𝑄₂ due to 𝑄₁?

04:49

Video Transcript

Point charges 𝑄 one equals 0.0050 nanocoulombs and 𝑄 two equals 0.020 nanocoulombs are fixed at π‘Ÿ one equals 9.00𝑖 plus 3.50𝑗 meters and π‘Ÿ two equals 6.50𝑖 minus 2.00𝑗 meters, respectively. What is the magnitude of the electric force on 𝑄 two due to 𝑄 one?

So we have these two charges: 𝑄 one and 𝑄 two. And let’s start out by drawing their position relative to one another. Working with this π‘₯, 𝑦-coordinate set, 𝑄 one is located at nine 𝑖 and 3.5𝑗, which puts that charge right there as we’ve drawn it. 𝑄 two on the other hand at 6.50𝑖 and negative 2.00 𝑗 is right there. And we can draw in the distance vectors π‘Ÿ one and π‘Ÿ two for each of these charges.

Now, in this question, we want to solve for the magnitude of the electric force on 𝑄 two due to 𝑄 one. And since we’re speaking of magnitude, we could equivalently solve for the force on 𝑄 one due to 𝑄 two; they’re the same. We can now recall Coulomb’s law, which tells us the electrical force that acts between two point charges 𝑄 one and 𝑄 two. It’s equal to the product of those charges times Coulomb’s constant π‘˜ all divided by the distance between the two charges squared.

Now, here’s a question: what’s the distance between 𝑄 one and 𝑄 two, our two charges? If we call that distance 𝑑, we know it’s equal to the magnitude of π‘Ÿ one minus π‘Ÿ two. If 𝑑 were a vector, it would look the way we’ve drawn it on our graph.

Here’s how the distance 𝑑 fits into the overall expression for the magnitude of the electric force on 𝑄 two. The magnitude of that force, which we’ll call simply 𝐹, is equal to Coulomb’s constant times the magnitude of 𝑄 one times 𝑄 two all over 𝑑 squared. Coulomb’s constant π‘˜ is approximately equal to 8.99 times 10 to the ninth newton meter squared per coulomb squared.

Knowing that as well as 𝑄 one, 𝑄 two and π‘Ÿ one and π‘Ÿ two, we’re ready to plug in and solve for the magnitude of 𝐹. What we’ve done here is we’ve plugged in our values for the numerator and left are denominator for now in terms of π‘Ÿ one and π‘Ÿ two.

Before addressing that denominator, let’s consider the way we’ve written our numerator. We see we’ve plugged in our value for Coulomb’s constant π‘˜. Then, we converted our charge 𝑄 one from the way it was expressed in the problem statement as 0.0050 nanocoulombs to a number in scientific notation, 5.00 times 10 to the negative 12th coulombs. We’ve done the same thing for our charge 𝑄 two, moving it from 0.020 nanocoulombs to 2.00 times 10 to the negative 11th coulombs. So that’s or upstairs.

But what about our downstairs, the denominator? We said that 𝑑, the distance between 𝑄 one and 𝑄 two, is equal to the magnitude of the vector π‘Ÿ one minus π‘Ÿ two. And since Coulomb’s law calls for this distance to be squared, we’ve squared this magnitude term. But how do we actually go about calculating the magnitude of π‘Ÿ one minus π‘Ÿ two when π‘Ÿ one and π‘Ÿ two are vectors?

We can say that if 𝑑 is equal to the magnitude of the difference between two vectors π‘Ÿ one and π‘Ÿ two, then that’s equal to the square root of the difference between the π‘₯-component of these vectors squared plus the difference between the 𝑦-component of the vectors squared. This is all assuming of course that π‘Ÿ one and π‘Ÿ two are two-dimensional vectors, as they are in our case.

We can use this expression in order to solve for magnitude of π‘Ÿ one minus π‘Ÿ two in our case. When we plug in for the given values of π‘Ÿ one and π‘Ÿ two by their components, we see that the magnitude of π‘Ÿ one minus π‘Ÿ two is equal to the square root of 9.00 minus 6.50 squared meter squared plus 3.50 plus 2.00 squared meter squared. Then, we notice that this entire expression in the square root sign is itself squared. So that means that the square and our square root sign cancel one another out. This means our denominator simplifies to 2.50 squared plus 5.50 squared meter squared.

Before we enter this expression on our calculator, take a look at the units. Notice that the units of meter squared in the numerator and in the denominator cancel out, as do the units of coulomb squared. So we’re left with units of newtons, which indicates we’re solving for a force. We are! So that’s a good sign that we’re on the right track.

Entering this whole expression on our calculator, to two significant figures, we find a result of 2.5 times 10 to the negative 14th newtons. That is the magnitude of the electric force on 𝑄 two due to 𝑄 one.

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