What is the Net Electric Field at Point P Due to Two Charges on the X-Axis?

[Hashmeer]

Homework Statement

Two charges Qa = 2 µC and Qb = -2 µC are placed on the x-axis with a separation of a = 21 cm.
(a) Find the net electric field at point P, a distance d = 13 cm to the left of charge Qa.

Homework Equations

E=F/q=kq/r^2

The Attempt at a Solution

I tried plugging into the above equation (each point separately) then summing them, but I just cannot seem to find the correct answer.

Thanks for the help.

 

[kuruman]

Hi Hashmeer, welcome to PF. Can you show us exactly what you did? Then we could pinpoint where you could have gone wrong.

 

[dacruick]

I think I can guess where you have gone wrong. Be sure to understand that you do not want to add them since they are on the opposite sides of the point of interest. This means that one charge will be pushing it away. and one charge will be pulling it towards. So both have to be added in the same direction.

 

[Hashmeer]

Okay so here is what I did:

Using the above equation, E=kq/r^2

kq(1)/r^2 + kq(2)/r^2 where q(1) = 2x10^-6 and q(2) = -2x10^-6.

Then the distance from P to q(1) is 13 cm and the distance from P to q(2) is 34 cm (the 21 cm between charges and the 13 cm from q(1) to P)

So it becomes kq(1)/(.13^2) + kq(2)/(.34^2)

Since the electric field for a negative points away and the positive is towards the signs are switched (that is the positive charge field is in the negative x direction and negative is positive direction).

I then get an answer of 7.73e6 which is wrong.



Also this is what the diagram looks like:

P----13cm----(q1)--------21cm---------(q2)

Hopefully this clears up the problem.

Thanks again.

 

[dacruick]

ohh that changes things. sorry about the misunderstanding. the answer i have is 9.1 x 10^-5. Your equation is right, so make sure that you know µC are 10^-6 and that k = 9x10^9.

 

[dacruick]

the only place i see you could have gone wrong is BEDMAS or your k value is off.