Question

A small sphere of mass m = 7.40 g and charge q1 = 31.2 nC is attached to the end of a string and hangs vertically as in the figure. A second charge of equal mass and charge q2 = −58.0 nC is located below the first charge a distance d = 2.00 cm below the first charge as in the figure.(a) Find the tension in the string. (b) If the string can withstand a maximum tension of 0.180 N, what is the smallest value d can have before the string breaks?

Answer 1

a) The hanging sphere experiences 3 forces, a gravity mg force pulling down, a tension pulling up, and an electric force, in this case opposites attract, so pulling it down. Therefore:

T = mg + Fe

The electrical force, Fe, is solved using coloumbs law, Fe = kq1q2/r²

Fe = (9e9) (31.2e-9) (58e-9 ) / (0.02 )²

Fe = 0.0407 N

Therefore,

T = mg + Fe

T = (0.0074)(9.8) + 0.0407

T = 0.113 N

b) We will treat Fe as the unknown, and T as the known

0.180 N = mg + Fe

0.180 N = (0.0074)(9.8) + Fe

Fe = 0.107

Now we use coloumbs law again, and solve for r

Fe = kq1q2/r²

0.107 = (9e9) (31.2e-9) (58e-9 ) / (r)²

r = 0.0123 m or 1.23 cm

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