Question

What is the direction of the force experienced by charge q in the diagram, if the other particles have charges of qa-qb +8.45 A/C, qc-qd--8.45 C, and q is positive? The shape is a square, and charge q is exactly in the center O up and left O down and right O up and right O down and left O left O down O up O right 4a 7b le Find the magnitude of the force on the charge q. The square has side length 10.60 cm, and the central particle's charge q 2.05 uC Number

Answer 1

From symmetry Force due to q_a & q_b cancels their horizontal component, while their vertical downward component of forces add up similarly, force due to q_c & q_d is vertically downward therefore Net force is downward Answer: down F_net = 4 times force due to each pair = 4 times kq q_a/(d/squareroot 2)^2 cos 45 degrees = 4 times 9 times 10^9 times 2.05 times 8.45 times 10^-12/squareroot 2 times (0.106/squareroot 2)^2 = 78.5 N Ans