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I need to find the work done by some stationary particle on another. The work is defined as a scalar product of force and displacement ($W=\int_{}^{} \vec{F} \cdot d \vec{r}$). The force is inversely proportional to the square of distance between the two particles: $F=k/r^2$ where $k>0$.

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Suppose the stationary particle is located at $r=a$. The other particle is located at $r=0$. Then the initial distance between them is $|\vec{r_0}|=a$. Later the other particle reaches some point where the distance between them is $r_1$. All the vectors here are in the direction of the $r$ axis, so they are all positive. Then the work done by the force $F$ is:

$$ W=\int_{r_0}^{r_1} \frac{k}{r^2} dr = -\frac{k}{r_1}-\left(-\frac{k}{r_0} \right) $$

But it turns to be wrong. The correct work should be $-\frac{k}{r_0}-\left(-\frac{k}{r_1} \right)$, that is the negative of my $W$. Where I messed up the signs?

user147535
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1 Answers1

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It's just that your $dr$ should be $-dr$ as you are decreasing $r$.

You can always check your answer whether it is positive or not. If Force and displacement are in same direction, work done by force must be positive...

evil999man
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