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, k_elec, 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.

r_TA [times] r_TA hat equals the position vector r_T, minus the position vector r_A, which is (x_T minus x_A) e_x plus (y_T minus y_A) 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 r_TA is 0.100 metres. And r_TA hat is e_x. Now you may have recognised this without needing to subtract the position vector r_A from r_T, but working through this in full is good practice for situations with more complicated geometry.

So if we substitute the magnitude r_TA and r_TA hat into Coulomb’s law, we get F_TA equals k_elec 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 F_TB. 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 F_TB must be F_TA rotated 90 degrees clockwise. So the vector F_TB equals the magnitude F_TA multiplied by minus e_y. And this means that F_TA plus F_TB 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 F_TB′ and we’ll add this to F_TA. Let’s work out our displacement vector from B′ to T. That’s the position vector r_T minus the position vector r_B′, 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 r_TB′. And we can work that out by drawing a right-angle triangle like this and then using Pythagoras’ theorem. So the magnitude r_TB 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 r_TB′ hat is the vector r_TB′ over the magnitude r_TB′.

So let’s substitute this expression for r_TB′ hat into Coulomb’s law. And we get F_TB prime equals k_elec times (30 times 10 to the minus 9 coulombs) squared, times the vector r_TB′, which is 0.100 metres (e_x minus e_y), divided by the magnitude r_TB′ 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 F_TA and F_TB′ 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.