# Transcript

[Text on screen: Using the vector form of Coulomb’s law to calculate electric forces on charged spheres]
SAM EDEN
SAM EDEN: In this screencast, we’re going to use the vector form of Coulomb’s law to calculate the forces between the charged spheres in our filmed experiments. So let’s start by sketching the experiment. We’ll assume that we can treat the charge spheres as point charges. So here’s our test charge T. And we’ll place it at the origin of a set of Cartesian coordinates. All our charges are on the xy-plane so we can treat this as a two-dimensional problem.
Charge A is at (minus 0.100 metres, 0.000 metres). I’m giving the coordinates to three significant figures in agreement with the experiment video. And the third charge can be at B, which is (0.000 metres, 0.100 metres). Or at B′, which is (minus 0.100 metres, 0.100 metres). Finally, our sensor measures F x, which is the x-component of any force that the test charge feels.
Now let’s write down the vector form of Coulomb’s law. The force on point charge 1 due to point charge 2 is written F12. And this is equal to a constant, kelec, which is 1 over (4πε0), multiplied by the charges q1 and q2, divided by the magnitude of the displacement vector r12 squared. And the direction of the force is given by the unit vector r12 hat.
Remember that r12 is the vector from point 2 to point 1. And we can express this as the magnitude r12 multiplied by the unit vector, r12 hat, which is equal to the position vector of point 1, r1, minus the position vector of point 2, r2.
OK, so our first situation has 30 nanocoulombs at T and at A. Let’s start by working out our displacement vector from A to T.
rTA [times] rTA hat equals the position vector rT, minus the position vector rA, which is (xT minus xA) e x plus (yT minus yA) e y.
Now we can substitute in our coordinates for T and A to give us (0 minus (minus 0.100 metres)) e x plus (0 minus 0.000 metres) e y, which is 0.100 metres times the unit vector e x. So the magnitude rTA is 0.100 metres. And rTA hat is e x. Now you may have recognised this without needing to subtract the position vector rA from rT, but working through this in full is good practice for situations with more complicated geometry.
So if we substitute the magnitude rTA and rTA hat into Coulomb’s law, we get FTA equals kelec times q1 q2, which is (30 times 10 to the minus 9 coulombs) squared, divided by (0.100 metres) squared, all multiplied by e x. And then crunching the numbers gives us 0.81 millinewtons e x.
Now on to our second situation. It’s the same as what we just looked at, except we’ve added a third sphere with 30 nanocoulombs at position B. Due to the principle of superposition of electric fields, the total force on T must be the sum of the forces due to the charges at A and B. So we need FTB. We can apply the method you just saw, but it’s quicker to work this out using symmetry.
If you rotate the charge at A 90 degrees clockwise about T, then it ends up at position B. So FTB must be FTA rotated 90 degrees clockwise. So the vector FTB equals the magnitude FTA multiplied by minus e y. And this means that FTA plus FTB equals 0.81 millinewtons e x, that we worked out just a moment ago, minus 0.81 millinewtons e y.
Now in our third situation, we have 30 nanocoulombs at T, A and B′. So we need the force FTB′ and we’ll add this to FTA. Let’s work out our displacement vector from B′ to T. That’s the position vector rT minus the position vector rB′, which is (0 minus (minus 0.100 metres)) e x plus (0 minus 0.100 metres) e y. And that equals 0.100 metres (e x minus e y).
Now we need the magnitude of rTB′. And we can work that out by drawing a right-angle triangle like this and then using Pythagoras’ theorem. So the magnitude rTB prime is the square root of (0.100 metres squared plus 0.100 metres squared), which is root (0.0200 metres squared). Now from the definition above we know that rTB′ hat is the vector rTB′ over the magnitude rTB′.
So let’s substitute this expression for rTB′ hat into Coulomb’s law. And we get FTB prime equals kelec times (30 times 10 to the minus 9 coulombs) squared, times the vector rTB′, which is 0.100 metres (e x minus e y), divided by the magnitude rTB′ cubed, which is (0.0200 metres squared) to the power of 3 over 2. And this comes to 0.29 millinewtons (e x minus e y).
Now we can add up FTA and FTB′ to get 0.81 millinewtons e x plus 0.29 millinewtons (e x minus e y), which is 1.1 millinewtons e x minus 0.29 millinewtons e y, keeping everything to two significant figures in line with the precision of our charge values. So our calculated F x is 1.1 millinewtons.
So now we’ve used vector components to calculate the force that acts on the test charge in each of the three situations in the experiment video.