Posted by **JJ** on Wednesday, July 18, 2012 at 7:52pm.

There are four charges, each with a magnitude of 2.65 microcoulombs. Two are positive and two are negative. The charges are fixed to the corners of a 0.408-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

I'm pretty confused, but this is what I think I should do for a start. If I were to draw a square and label the charges at the 4 corners q1, q2, q3, and q4 starting at the top left corner and numbering around clockwise; then put q1 and q4 as negative, and q2 and q3 as positive. This would direct the force inward, I believe? Then I think another force on each would point away from the square? So, if I were to pick to solve for q2, then would it be F21 + F31 for the force on q2? I'm also unsure if I have to consider the second force on each charge and use cos and sin for the x and y axis? Help would be very appreciated.

- Physics -
**bobpursley**, Wednesday, July 18, 2012 at 8:17pm
let the side distance be s. Then the diagonal distance is s*sqrt2

Now each charge is attracted to the side charges, and repelled by the opposite charge across the diagonal.

Now consider the charge at a corner being attracted to the two adjacent charges. The force of attraction is kqq/s^2, and the directions of these two forces are perpendicular to each other, so the resultant force is along the diagonal, and has a value of 2*.707*kqq/s^2 in the direction of the diagonal toward the center. Now consider the force across the diagonal due to the charge across the diagonal

force= kqq/(ssqrt2)^2=1/2 *kqq/s^2, but it is being repelled.

Net force toward the center:

kqq/s^2*1.414-1/2 kqq/s^2=about you do it.

So each charge is attracted towards the center. OH yes, this is for the charges to be alternated in +- going around the square.

- Physics -
**JJ**, Wednesday, July 18, 2012 at 9:43pm
Where is the 0.707 coming from? And I plugged k= 8.99 x 10^9, q= 2.65 x 10^-6, and s= 0.408 into the net force toward the center equation to get 0.347N. So this is the magnitude that all of the charges would have, correct?

## Answer This Question

## Related Questions

- college physics - There are four charges, each with a magnitude of 2.0uC. Two ...
- physics - Four point charges have the same magnitude of 2.62 10-12 C and are ...
- Physics - Three Point charges have equal magnitudes, two being positive and one ...
- Physics - Three Point charges have equal magnitudes, two being positive and one ...
- Physics - 1. The problem statement, all variables and given/known data A point ...
- physics - Four point charges are located at the corners of a square with sides ...
- Physics - Three point charges have equal magnitudes, two being positive and one...
- Physics - Three point charges have equal magnitudes, two being positive and one...
- Physics - 1.There are four charges, each with a magnitude of 2.5 µC. Two are ...
- Physics - 1) Charges of +4uC and -6uC are placed at two corners of an ...

More Related Questions