Question

A charge of 6.00 mC is placed at each corner of a square 0.480 m on a side. Determine the magnitude of the force on each charge. Determine the direction of the force on each charge.

Answer 1

The force exerted on Charge, #2 by Charge # 1,
F21 = k.q.q/a2 =9x109x6x6x10-6/0.1x0.1
= 3.27x107 N/C along the x axis
Similarly,
F23 = 3.27x107 N/C alongthe y axis
And F24 =k.q.q/2a2
= 3.27x107 x1/2 N/C along the diagonaloutward
The resultant of the first two forcesis F21 x2/v2= F21 xv2 along the diagonal, samedirection as F24 .
therefore the total of all the forces on #2 is,
F2 = F21x(0.5 +1.414) = 1.914 F21
= 1.914 x 3.27x107 = 6.20 x107 N/C
along the diagonal outward; at 45o with thesides.
Similar for all other charges.

Answer 2

F-Q102 4TT2E0

q1 = q2 = 6mc = 0.006 c

r =0.480√2 = 0.68m

(1/4πεo) = 8.98*10⁹

=> F = 323.28 kN

Direction will be along the diagonal of the square

Answer 3

First find the force of the two particles that are not on a diagonal. Since each particle produces and equal force, the charge will be propelled diagonally, at an angle of 45 degrees.

F = kq1q2/r2 = (8.99 * 10^9)(6*10^-3)2 / (0.48)2 = 1406250 N

1406250 + 1406250 = c2

c = 2812500 N

Now, we solve for the third charge on the diagonal.

F = kq1q2/r2 = (8.99*10^9)(6*10^-3)2 / 0.4608

F = 703125 N

Now just add the two and the answer is yours.

F = 703125 + 2812500 = 3515625 N = 3.5 * 10^6 N