Posted by **Jessica** on Wednesday, February 20, 2008 at 1:32am.

There are two +q charges and two -q charges on the corners of a square of size a. Left upper corner= +q charge, right upper corner= -q charge, left lower corner= -q charge, and right lower corner= +q charge. Compute the net force on each charge.

I know that F= 9x10^9 Nm^2/C^2 x qaxqb/r^2 and that all the charges are attracted to eachother. so would i just find the forces for each charge using that equation, but how do i deal with it pulling in different directions?

- physics -
**drwls**, Wednesday, February 20, 2008 at 2:17am
For each charge, you have to add the forces due to the three other charges as vectors acting in different directions. Some of the forces are attraction and some are repulsion.

You have the correct formula for the force on each pair of charges.

When you add the forces due to the two adjacent corners, it will be an attraction actng along the diagonal, sqrt2 times larger than the force due to either adjacent corner. The opposite corner will generate a repulsion force in the opposite direction, but lower than the combined effect of the other two forces.

## Answer This Question

## Related Questions

- Physics - Three charges are located at the corners of a rectangle as follows: ...
- Physics - Three charged particles are arranged on corners of a square as shown ...
- Physics Superpostion Principle - four charged particles are placed so that each ...
- Physics - On the lower left corner of a square is a fixed charge qLL = 1.5x10^-9...
- Physics 1112 Electric charge - As shown in the figure, a square has sides of 8.0...
- Physics - A rectangle has a length of 2d and a height of d. Each of the ...
- Physics - In the rectangle in the drawing, a charge is to be placed at the empty...
- physics - A rectangle has a length of 2d and a height of d. Each of the ...
- Physics - At each corner of a square of side script i there are point charges of...
- Physics - I don't really know where to start on the problem. Would I use V = k(q...

More Related Questions