Question

Problem 3 A charge Q is placed at the centre of the square of side 2.10 cm, at the corners of which four identical charges q-7.5 C are placed. Find the value of the charge Q so that the whole system is in equilibrium cmtacd n te value of the charge Oa! 7.98 What is the condition on the net force on a charge for it to be in equilibrium? Subonit AnswerIncorrect. Tries 1/6 Previous Trics

Answer 1

Individual forces.

Forces from the adjacent corners sum to
F1 = 2k(q/s)²/sqrt(2) = sqrt(2)k(q/s)²
Force from the opposite corner, where distance = sqrt(2)s,
F2 = k(q/(sqrt(2)s))² = 0.5k(q/s)²
Total force F = F1+F2 = (sqrt(2)+0.5)k(q/s)²
Now just calculate -F = kQq/r² ==> Q = -Fr²/(kq), where r = s/sqrt(2), or
Q = -Fs²/(2kq) = (sqrt(2)+0.5)q/2.
Note this is an unstable equilibrium.
For the values given, I get Q = -7.178 C