# A bit more physics help

• Sep 16th 2006, 10:39 AM
Jones
A bit more physics help
Hi, again...
I have another problem im unable to solve =,(
The task is to calculate the elecrical field in the point P. I do know we need to use coloumbs law
but i don't how the force from the two charged particles affects P.
An image i made to make it more legible :o

• Sep 16th 2006, 11:01 AM
topsquark
Quote:

Originally Posted by Jones
Hi, again...
I have another problem im unable to solve =,(
The task is to calculate the elecrical field in the point P. I do know we need to use coloumbs law
but i don't how the force from the two charged particles affects P.
An image i made to make it more legible :o

The formula for the electric field is:
E = (kq)/r^2
Where k is the Coulomb constant 9.0 x 10^9 Nm^2/C^2, q is the charge at some point away, and r is the distance between the charge and the observation point.

Now, E is a vector, so what you need to do is determine the magnitude and direction of E due to each charge, then add the E values vectorally. The rule for assigning the direction to E is simple: The E field lines always radiate outward from a positive charge, and radiate inward to a negative charge. (So E due to the charge at point A is downward, and E due to the charge at B is to the left.)

I'll leave it here for now. If you have problems adding the vectors (most students have forgotten most of what they know about adding vectors by the time they start electric fields) just let me know.

-Dan
• Sep 16th 2006, 12:50 PM
Jones
Quote:

Originally Posted by topsquark
The formula for the electric field is:
E = (kq)/r^2
Where k is the Coulomb constant 9.0 x 10^9 Nm^2/C^2, q is the charge at some point away, and r is the distance between the charge and the observation point.

Now, E is a vector, so what you need to do is determine the magnitude and direction of E due to each charge, then add the E values vectorally. The rule for assigning the direction to E is simple: The E field lines always radiate outward from a positive charge, and radiate inward to a negative charge. (So E due to the charge at point A is downward, and E due to the charge at B is to the left.)

I'll leave it here for now. If you have problems adding the vectors (most students have forgotten most of what they know about adding vectors by the time they start electric fields) just let me know.

-Dan

And that would give us something like this:http://aycu16.webshots.com/image/289...4110982_rs.jpg

But how long should E1 and E2 be?
• Sep 16th 2006, 03:27 PM
topsquark
Quote:

Originally Posted by Jones
And that would give us something like this:http://aycu16.webshots.com/image/289...4110982_rs.jpg

But how long should E1 and E2 be?

E1 = k(q1)/(r1)^2 and Ex = k(q2)/(r2)^2.

And E1 is in the direction of your "other" left. :) It's a negative charge so the field lines point toward it, not away from it.

Specifically: E1 = (9.0 x 10^9)(36 x 10^(-6))/(6.0 x 10^)-2))^2 N/C
E2 = etc.

-Dan
• Sep 16th 2006, 10:33 PM
Jones
Quote:

Originally Posted by topsquark
E1 = k(q1)/(r1)^2 and Ex = k(q2)/(r2)^2.

And E1 is in the direction of your "other" left. :) It's a negative charge so the field lines point toward it, not away from it.
-Dan

I know that, just wanted to make sure you knew....

;)