1
$\begingroup$

Let's say we have two charges called $q_1$ and $q_2$, respectively $20 \, C$ and $-40\,C$, at a distance $d=1\,m$ We want to find all the points where electric potential is null.

I solved the equation $$\frac{q_1}{4\pi\epsilon_0r_1} + \frac{q_2}{4\pi\epsilon_0(d-r_1)}=0$$ For $r_1$ (distance from $q_1$), and found $r_1=\frac13\,m$

However this is not the only solution: there is another point not in-between the charges, but $1\,m$ left from $q_1$ ($r_1=-1\,m$)

How can I set up an equation giving me both the solutions?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

Let's draw the setup:

Charges

The expression for $V(r)$ is simply (I'll set $\kappa$ to 1 for convenience):

$$ V(r) = \frac{20}{r} - \frac{40}{r+1} $$

so the potential is zero when:

$$ \frac{20}{r} = \frac{40}{r+1} $$

Only this isn't quite right because the potential for each charge is symmetric so the potential due to charge $A$ obeys $V_A(r) = V_A(-r)$ and likewise for the other charge. It's because you're ignoring this that your equation gives you only one null point. The equation really should be:

$$ \frac{20}{|r|} = \frac{40}{|r+1|} $$

The easy way to deal with those modulus operators is to square both sides:

$$ \frac{400}{r^2} = \frac{1600}{(r+1)^2} $$

and if we rearrange this we get the quadratic equation:

$$ 3r^2 - 2r - 1 = 0 $$

Quadratic equations have two roots, and the two roots are going to give you the two null points. If we use the usual expression for the roots of a quadratic equation we get $r = 1$ and $r = -\tfrac{1}{3}$.

$\endgroup$
5
  • $\begingroup$ I suspected we were searching for some squares, but I didn't think of square my equation because I thought I was lacking a more fundamental step $\endgroup$
    – seldon
    May 17, 2014 at 7:54
  • $\begingroup$ We're all so used to writing $V \propto 1/r$ that it's very easy to forget it's actually $V \propto 1/|r|$ :-) $\endgroup$ May 17, 2014 at 7:59
  • $\begingroup$ I think it's because $r$ is considered geometrically (thus always positive), while now we talk about it as a signed number, a point in a Cartesian plane. Does it make sense? $\endgroup$
    – seldon
    May 17, 2014 at 8:54
  • $\begingroup$ @mattecapu Yes, you are right. $\endgroup$
    – evil999man
    May 17, 2014 at 9:03
  • $\begingroup$ However, thank you very much for your detailed answer! I really appreciated it $\endgroup$
    – seldon
    May 17, 2014 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.