I'm having trouble knowing how to go about solving this question:

Q: The force on a particle at a point with position vector $r = xi + yj + zk$ exerted by a charge at the origin is $F(r)=\left(\frac{P(r)}{|r|^2}\right)$ in which P is constant. Calculate the work done as the particle moves in a straight line from (1, 0, 0) to (1, 2, 3).

What I think I need to do:
$r_1=i$, $\quad$ $r_2=i+2j+3k$ $\quad$ so let $r(t)=i+2tj+3tk$,$\quad$ $0\lt t \lt 1$
Then $\frac{dr}{dt}=2+3=5$ and $|r|^2=1+13t^2$
Therefore, $$F(r)=\frac{P(i+2tj+3tk)}{1+13t^2}$$ As such, $\int_C F(r) \cdot dr=\int_0^1 \frac{P(i+2tj+3tk)}{1+13t^2} \cdot5 dt=\int_0^1 \frac{5P(1+5t)}{1+13t^2}dt=5P\int_0^1 \frac{1+5t}{1+13t^2}dt$

Is this the right way to go about answering this question, or am I doing something completely wrong? Any advice would be greatly appreciated.

  • $\begingroup$ Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector. $\endgroup$ – PierreCarre Feb 1 at 13:26

The work is computed as the line integral:

$$ \int_C F(r)\cdot dr = \int_0^1 F(r(t))\cdot r'(t) dt = \int_0^1 \frac{P}{1+4t^2+9t^2} (1,2t,3t)\cdot(0,2,3) dt = \int_0^1 \frac{13 P \,t}{1+13t^2}dt $$

  • $\begingroup$ Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral? $\endgroup$ – Anon Feb 1 at 13:58
  • $\begingroup$ Gone with the wind... I'll edit the post! $\endgroup$ – PierreCarre Feb 1 at 16:06

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