They tell you what the charges are. There are three +2nC charges and one -2nC charge.
They ask you for the FIELD at the center of the square, not the force. There is no point charge there.
Posted by jude on Saturday, February 6, 2010 at 2:20am.
Given a square of sidelength a = 5 cm. We place a charged particle at each corner, three of them carry + 2 nC of charge and one carries the same amount of negative charge.
What is the magnitude of the electric field at the center of the square?
I don't know how to start this problem and would really appreciate it if someone could tell me the method.
Thank you
* physics - drwls, Saturday, February 6, 2010 at 2:27am
The center of the square is 5/sqrt2 = 3.535 cm from each corner. The fields due to the two +2nC charges at opposite corners will cancel out. The other two opposite charges will add, and the field will act along the diagonal between them.
Calculate the E field due to the negative charge using Coulomb's Law, and double it to account for the positive charge at the opposite corner.
* physics - Sandra, Saturday, February 6, 2010 at 2:25pm
Is the point charge considered to have a charge of 1?
6 answers
Thank you for clearing that up!
I'm getting a wrong answer once again for this question though.:( I tried it your way, but I have a feeling I'm making a mistake somewhere. Please help!
Here's what I did:
For the -ve charge located at the top right coner of the square: there are 3 forces acting on it. 1 pointing left, 1 pointing down and one pointing diagonally towards the center of the square.
I found each of these 3 forces; where the one pointing left (F1) and up(F3) are of equal magnitude and F2 is the one one that is pointing .
F1 = F3 = k(2*10^-9)(2*10^-9) / (0.04)^2 = 2.248E-5
F2 = k(2*10^-9)(2*10^-9) / (0.05657)^2 = 1.1239E-5
(Note: 0.05657 = sqrt(0.04^2 + 0.04^2) = the distance between the diagonally located +ve and -ve charges)
Since F2 was a diagonal force, I found its components F2x and F2y:
F2x = F2cos45 = 7.947E-6
F2y = F2sin45 = 7.947E-6
Now the net force on the -ve charge was found :
Fnet = sqrt( (F1+F2x)^2 + (F3+F2y)^2 ) = 4.303E-5
E = F/q = 4.303E-5 / 2E-9 = 21514.48
You said this electric field should be doubled,
so my final answer for this question was:
Etot = 43028.96 N/C or 43.0 kN/C
I'm getting a wrong answer once again for this question though.:( I tried it your way, but I have a feeling I'm making a mistake somewhere. Please help!
Here's what I did:
For the -ve charge located at the top right coner of the square: there are 3 forces acting on it. 1 pointing left, 1 pointing down and one pointing diagonally towards the center of the square.
I found each of these 3 forces; where the one pointing left (F1) and up(F3) are of equal magnitude and F2 is the one one that is pointing .
F1 = F3 = k(2*10^-9)(2*10^-9) / (0.04)^2 = 2.248E-5
F2 = k(2*10^-9)(2*10^-9) / (0.05657)^2 = 1.1239E-5
(Note: 0.05657 = sqrt(0.04^2 + 0.04^2) = the distance between the diagonally located +ve and -ve charges)
Since F2 was a diagonal force, I found its components F2x and F2y:
F2x = F2cos45 = 7.947E-6
F2y = F2sin45 = 7.947E-6
Now the net force on the -ve charge was found :
Fnet = sqrt( (F1+F2x)^2 + (F3+F2y)^2 ) = 4.303E-5
E = F/q = 4.303E-5 / 2E-9 = 21514.48
You said this electric field should be doubled,
so my final answer for this question was:
Etot = 43028.96 N/C or 43.0 kN/C
This answer was wrong though when I submitted it. If it's possible, can you please direct me to the mistake I'm making?
Only two charges contribute to the field at the center as WLS said. The two positive charges that are not opposite the negative charge cancel each other at the center of the square. Read WLS instructions again.
Also you are not interested on forces on the negative charge. You are interested in the field at the center of the square.
ok..so how would we find the forces acting at the center of the field? There will be two forces acting, one form the negative and one from the positive charge right? But if we use Coulomb's Law here.. wouldn't we need a charge in the center? I don't understand how Coulomb's law i.e Fe = kq1q2/r^2 can be applied here...
1 answer