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Question about force between particles

I'm trying to work out the force between a proton and an electron. I know I'm supposed to use the formula F = kQ(1)Q(2) / r^2.
k is given as 9.0 x 10^9 m/F and the distance between the particles is 5.0 x 10^-8 m.

I have the charge for the electron (1.6 x 10^-19 c), but what about the proton? There is an example in the book and the charge for Q(2) the charge is the same as Q(1), but I don't get the answer even if I do that. How do I do it?
Reply 1
I will show my working so far

F = (9.0 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (5.0 x 10^-8 x 5.0 x 10^-8)

Where did I go wrong?
Force between two charged particles, one of charge Q1 Q_1 and one of charge Q2 Q_2, separated by a distance r in the vacuum is given by coloumbs law:

F=14πϵ0Q1Q2r2 F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2}

So for a proton and electron separated by 5×108m5 \times 10^{-8} m, the force is: F=14πϵ0(1.6×1019C)2(5×108m)2 F = \frac{1}{4 \pi \epsilon_0} \frac{(1.6 \times 10^{-19} C)^2}{(5\times 10^{-8}m)^2}

Should be: |F| = 9.2 x 10^{-14} N
Reply 3
Chaotic Evil
I will show my working so far

F = (9.0 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (5.0 x 10^-8 x 5.0 x 10^-8)

Where did I go wrong?


I dont think you have gone wrong. Looks Ok to me.
Reply 4
0 div curl F
Force between two charged particles, one of charge Q1 Q_1 and one of charge Q2 Q_2, separated by a distance r in the vacuum is given by coloumbs law:

F=14πϵ0Q1Q2r2 F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2}

So for a proton and electron separated by 5×108m5 \times 10^{-8} m, the force is: F=14πϵ0(1.6×1019C)2(5×108m)2F = \frac{1}{4 \pi \epsilon_0} \frac{(1.6 \times 10^{-19} C)^2}{(5\times 10^{-8}m)^2}

Should be: |F| = 9.2 x 10^{-14} N


That's what I got, but instead of multiplying the whole thing by 14πϵ0 \frac{1}{4 \pi \epsilon_0} they've multiplied the numerator by it (9 x 10^-9) and the denominator by 8.85 x 10^-12, which is epsilon 0, to get the answer.

Thanks both of you, I just needed to understand how to do it, I tried a different questions this way and it was right.

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