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## Homework Statement

"Two tiny chips of plastic masses 4.5x10^-5 are separated by a distance of 1.5 mm. Suppose that they carry equal and opposite electrostatic charges +/-q. What must the magnitude of the charge be if the electric attraction between them is equal to the weight of one chip?"

## Homework Equations

Fe=k*[(qxq)/d^2]

weight= mass (in kg) x 9.8 m/s^2

q=charge

d=distance

k=Coloumb's constant, 8.99x10^9

## The Attempt at a Solution

1. First I have calculated the weight of one mass: .0045kg x 9.8 = .0441 N

2. then I set that equal to Coloumb's eq to try to solve for q. I divided k to the left so .0441 / 8.99x10^9 = 4.9054x10^-12.

3. I multiplied 4.9054x10^-12 by d^2 (or .0015^2) and got 1.1x10^-17

4. At this point I have q^2=1.1x10^-17 so I do a square root and get 3.322x10^-9. I have tried to tweek my answer so many times but to no more avail. I have spent almost an hour on this problem (pathetic, I know) but this problem set is due tomorrow and smartwork is being uncooperative. I would really appreciate any help!