We know that the force generated by an electric charge that is located at the origin, on a charged particle at a point $(x,y,z)$ of position vector $\vec{r}=x\space\vec{i}+y\space\vec{j}+z\space\vec{k}$ is $\vec{F}(\vec{r})\space=\space K\frac{\vec{r}}{\|\mathbf{{\vec{r}} \|}^3}$
I try to find the work of $\vec{F}$ when the particle is moving on a line from the point $(2,0,0)$ to the point $(2,1,5)$.
I am first starting by finding the parametric equation of the line. I have $$ \vec{r}(t)=2t\vec{i}+t\vec{j}+5t\vec{k} \\0\le t\le 1 $$
Now, I am looking for the magnitude of the line. I find
$$ \|\mathbf{\vec{r}}\|= t\space\sqrt{30} $$
After finding $\vec{r}'(t)$ and $\vec{F}(\vec{r}(t))$, I use the equation
$$ \int_0^1 \vec{F} \cdot d\vec{r}\ $$
And I have as result
$$ \frac{1}{30\sqrt{30}}\int_0^1 \frac{1}{t^2}(2\vec{i}+\vec{j}+5\vec{k}) $$
Which is impossible knowing that the inferior extreme is 0.
Am I making a mistake in the steps I use or in my calculations ?
Thank you.
An alternative approach: The force field is conservative, so the line integral is independent of the path. Instead of moving directly along the line joining the two points, you can instead move the particle directly away from the origin until you get to $(\sqrt{30},0,0)$ and then along the equipotential line $r=\sqrt{30}$ until you get to the endpoint. The second segment contributes nothing to the integral ($\vec F\cdot d\vec r=0$), while the first is simply $$K\int_2^\sqrt{30}\frac1{x^2}\,dx=K\left(\frac12-\frac1{\sqrt{30}}\right).$$ I’ll leave it to you to verify that this is the same value that you get by evaluating $\int_0^1\frac K{\|\vec r(t)\|^3}\vec r(t)\cdot\mathrm d\vec r(t)$ with $r(t)=2\vec i+t\vec j+5t\vec k$.