Thread: Coulombs law and Electric Fields

1. Coulombs law and Electric Fields

HI there. I am revising for my A level Physics exam in January. I have printed out past paper questions, but have no qnswers to them! I have worked through this question and was wandering if anyone could check to see I have done it correctly.

Any help greatly appreciated.

Two point charges, each of magnitude 5x10^-6C, are placed horizontally 50mm apart at points A and B respectively.

Calculate the force exerted on a point charge of 3x10^-6C at a point P, which is 20mm from A and 30mm from B, in the cases when:

i) both charges A and B are positive (my answer 11.25N)
ii) A is positive and the charge at B is negative (my answer 2.25N)

Thank you very much for assistance!

Mike

2. Originally Posted by mikewhant
HI there. I am revising for my A level Physics exam in January. I have printed out past paper questions, but have no qnswers to them! I have worked through this question and was wandering if anyone could check to see I have done it correctly.

Any help greatly appreciated.

Two point charges, each of magnitude 5x10^-6C, are placed horizontally 50mm apart at points A and B respectively.

Calculate the force exerted on a point charge of 3x10^-6C at a point P, which is 200mm from A and 30mm from B, in the cases when:

i) both charges A and B are positive (my answer 11.25N)
ii) A is positive and the charge at B is negative (my answer 2.25N)

Thank you very much for assistance!

Mike
if P is 30mm from B, and A and B are 50mm apart, how can P be 200mm from A?

did you mean to type 20mm?

3. sorry, yes, it is 20mm! thanks for pointing that out!

4. I'm afraid my calculations differ from yours.

How did you attempt to find the net force on charge P ?

5. Ok, well I worked out the electric field strength due to A (2.25x10^6N/C) and then the same for B (1.5x10^6N/C)

Then I worked out the force due to A (6.75N) and then for B (4.5N) and added them together giving 11.25N for part i

For part ii, I said that since B was negative, it should be subtracted from A giving 2.25N for part ii.

Mike

6. Originally Posted by mikewhant
Ok, well I worked out the electric field strength due to A (2.25x10^6N/C) and then the same for B (1.5x10^6N/C)

Then I worked out the force due to A (6.75N) and then for B (4.5N) and added them together giving 11.25N for part i

For part ii, I said that since B was negative, it should be subtracted from A giving 2.25N for part ii.

Mike
coulomb constant ... $k = 8.99 \times 10^9$

$q_A = 5 \times 10^{-6}$ C

$q_P = 3 \times 10^{-6}$ C

$r = 2 \times 10^{-2}$ m

force exerted on charge P by charge A ...

$F = \frac{kq_Aq_P}{r^2} \approx 337$ N

7. Got it!

Not too sure what I was trying to work out! I was using the equation

E=kq/r^2 to work out the elctric field strength. Didnt need to do that!

Thanks alot for that, take care

Mike