# Coulombs law and Electric Fields

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• Dec 30th 2009, 07:08 AM
mikewhant
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
• Dec 30th 2009, 07:15 AM
skeeter
Quote:

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?
• Dec 30th 2009, 07:19 AM
mikewhant
sorry, yes, it is 20mm! thanks for pointing that out!
• Dec 30th 2009, 07:31 AM
skeeter
I'm afraid my calculations differ from yours.

How did you attempt to find the net force on charge P ?
• Dec 30th 2009, 07:39 AM
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.

Thanks again for your help.

Mike
• Dec 30th 2009, 07:50 AM
skeeter
Quote:

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.

Thanks again for your help.

Mike

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

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

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

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

force exerted on charge P by charge A ...

$\displaystyle F = \frac{kq_Aq_P}{r^2} \approx 337$ N
• Dec 30th 2009, 07:55 AM
mikewhant
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