Wednesday

March 4, 2015

March 4, 2015

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

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 8:17pmlet 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 9:43pmWhere 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?

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