Three balls, with charges of +4q, -2q, and -q, are equally spaced along a line.

Question

Three balls, with charges of +4q, -2q, and -q, are equally spaced along a line. The spacing between neighboring balls is r. We can arrange the balls in three different ways, as shown in the figure. In each case, the balls are in an isolated region of space very far from anything else. +40 +4q (a) Rank the three cases, from largest to smallest, based on the magnitude of the net force experienced by the ball with the +4q charge. (Use only">" or symbols. Do not include any parentheses around the letters or symbols.) 1>3 2 (b) If q is 6.20 × 10-6 c and r is 30.0 cm, calculate the magnitude and direction of the net force acting on the ball with the +4q charge in case 1, (Define the positive direction as pointing to the right, so your answer should have a positive sign if the net force is directed right, and a negative sign if the net force is directed left.) 30.78

Explanation / Answer

in case1 :

due to -2q

F1 = k q1 q2 / r^2 = k (4q)(2q)/r^2 = 8 k q^2 / r^2

to the right  

F2 = k (4q)(q)/(r+r)^2 = k q^2 / r^2

to the right  

Net force on 4q:

F = F1 + F2 = 9 k q^2 / r^2

= (9)(9 x 10^9)(6.20 x 10^-6)^2 / (0.30)^2

= 34.6 N ......Ans

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