The force between two particles, both of charge "Q", is "F" when they are a distance "r" apart. Explain what the values of the force will be when:
d) The charges are Q^(1/2) and Q^(1/2) and the separation is r^(1/2)
Equation needed: F=(Q₁*Q₂)*(4πε0r² )⁻¹
ε0 = permittivity of free space
(4πε0)⁻¹, since this is constant it wouldn't affect the New force ( I will refer to this as k)
For this question both Q₁ and Q₂ is the same so i will just call this Q²
So rewriting the equation:
My answer for this question would be that the New F is (r/Q)*(Old F), however my teacher and the answers for the text book both say it is New F = Old F; although given an explanation it just doesn't make any logic sense to me.
This is what i have in mind: since k is constant i can just rearrange so F/k=(Q²/r²) for the original equation and F/k= (Q/r) for the new equation, from my understanding of what my teacher is telling me, he is saying (Q²/r²) = (Q/r), which is incorrect.
Others in my class also agrees with my teacher, maybe i've just missed something.
Someone please explain so i can put my mind at ease and sleep tonight.
Question 1d of Page 154 in OCR A Level Physics A 2015: Student book 2 (OCR GCE Science 2015)
I appolgise for any poor grammar or structuring.
Pus i am also happy to give addition information, maybe.
Electric fields; Coulomb's law Watch
- Thread Starter
- 27-09-2016 14:29