A positive charge of 3.0e-6 C is pulled on by two negative charges. One, -1.9e-6 C, is 0.045 m to the north and the other, -3.7e-6 C, is 0.030 m to the south. What total force is exerted on the positive charge?
F=kqq/d^2
You would have to find the F exerted by each. The first negative charge is going to attract to the positive charge and the second negative charge is going to attract to the negative charge. Does this have anything to do with way in which you add up the F in order to find the F exerted on the positive charge?
2 answers
I don't believe the negative charges will be attracted to each other. Yes, it has something to do with the way you add up the F but note that one negative charge is north, the other one is south; therefore, they are exactly opposite each other.
When I plug the numbers I have into the equations I get this:
F=(9e9)(-1.9e-6)(3e-6)/(.045^2)=-25.3 N
F=(9e9)(-3.7e-6)(3e-6)/(.030^2)=-111 N
Is this correct? If it is, in order to get the total force exerted on the positive charge, would I add this two up or subtract them from one another?
F=(9e9)(-1.9e-6)(3e-6)/(.045^2)=-25.3 N
F=(9e9)(-3.7e-6)(3e-6)/(.030^2)=-111 N
Is this correct? If it is, in order to get the total force exerted on the positive charge, would I add this two up or subtract them from one another?
2 answers