## Math-related Physics question.

1. The problem statement, all variables and given/known data
The electric charge of an electron is -1.6 x 10(to the -19th power) C. What is the force exerted between two electrons separated by one meter? (Be sure to note whether the force is attractive or repulsive.) Show your calculations.

2. Relevant equations
This is the equation I've been trying to use to solve it.
F= K q1q2/d squared

3. The attempt at a solution
I set up the equation and everything but I couldn't figure out the answer. The powers are confusing me. And I know that it is going to be labeled in Newtons but I don't understand how.

9.0 x 10(to the 9th power) N x m(squared)/c(squared) is what I used for K. Is that right?

Any help would be great. Thanks.

-natalia
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 Mentor The magnitude of the force between two charged particles is $$F=k\frac{|q_1||q_2|}{d^2}$$. Your value of k is correct. So, plugging in the values for the charge of an electron into this formula for q_1 and q_2 should yield the result. Now, to answer your question re: units. K has units N m2C-2, so let's plug the units into the equation $$[F]=\frac{Nm^2}{C^2}\cdot\frac{C\cdot C}{m^2}=N$$ and thus we see that the units of F are indeed Newtons, as required. Finally, intuition should tell you whether the force is repulsive or attractive. If two particles both have negative charge, will they attract or repel one another?

 Quote by cristo The magnitude of the force between two charged particles is $$F=k\frac{|q_1||q_2|}{d^2}$$. Your value of k is correct. So, plugging in the values for the charge of an electron into this formula for q_1 and q_2 should yield the result. Now, to answer your question re: units. K has units N m2C-2, so let's plug the units into the equation $$[F]=\frac{Nm^2}{C^2}\cdot\frac{C\cdot C}{m^2}=N$$ and thus we see that the units of F are indeed Newtons, as required. Finally, intuition should tell you whether the force is repulsive or attractive. If two particles both have negative charge, will they attract or repel one another?

Thank you. This helps a lot.

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