If I’m working in two dimensions and I wanted to calculate the net electrostatic force on charge C, where

  • charges A and B are stationary, of the same charge, and placed on the y-axis equidistant from the origin,
  • charge C is placed on the x-axis,
  • all charges are like charges with charge magnitude $Q$

I need only summate the x-components of the forces $F_{AC}$ and $F_{BC}$ since the y-components cancel wherever charge C is on the x-axis. If $\theta$ is the angle between the line extending from A to C ($r$) and the x-axis, then $$F_{net}=\frac{2kQ^2}{r^2}\cos\theta\tag1$$

That’s pretty simple and what we’re used to doing. However, my question is why can’t we take the x-component of $r$ first, and then use that in Coulomb’s law to find the x-component of the force? That is, $$r_x=r\cos\theta$$ $$F_{net}=\frac{2kQ^2}{r_x^2}=\frac{2kQ^2}{(r\cos\theta)^2}\tag2$$

The fact that this is a square relationship may change things, but even without it, equation 1 ≠ equation 2. This question extends to all general cases of resolving vectors. What’s an intuitive understanding of why you can’t do it the second method?

  • 3
    $\begingroup$ There is no reason (2) should work... why do you think it should? You're not the only person... I've seen other students over the years have the same expectation, but I'm not sure why. The net force between two charged particles is given by the coulomb formula. Then you can get components of that force using trigonometry. You have an "intuition" that (2) should work, but many times our intuitions are misleading. $\endgroup$ – Ameet Sharma Feb 13 '16 at 4:34

While your initial symmetry-based shortcuts are correct, they are hiding the answer to your question. Let's start at the beginning. The net force will be $$\vec{F}_{net}=\Sigma\frac{kQ^2\hat{r_i}}{r_i^2}.$$

$$\hat{r}_i=\frac{\vec{r_i}}{r_i}\text{ and }\vec{r_i}=x_C\hat{i}+(-y_i)\hat{j}\text{ and } r_i=\sqrt{x_C^2+y_i^2}$$

Because $y_A=-y_B=Y$ by setup of the problem, $r_{A}=r_B = \sqrt{x_C^2+Y^2}$. This is the total distance from each charge $A$ and $B$ to charge $C$. So the denominator in each term of the sum is identical. Let's call that distance $R$.

$$\vec{F}_{net}=\frac{kQ^2}{R^2}\frac{\left(x_C\hat{i}+(-Y)\hat{j}+x_C\hat{i}+(Y)\hat{j} \right)}{R},$$

which reduces to $$\vec{F}_{net}=\frac{kQ^2}{R^3}2\left(x_C\hat{i} \right).$$ With your definition of $\theta$, $x_C=\pm R\cos\theta$, so

$$\vec{F}_{net}=\frac{2kQ^2}{R^2}\cos\theta\left(\pm\hat{i} \right), \text{depending where C is positioned}.$$

The distance term in the denominator must contain the total distance to the charge, while you can split the vector portion of the numerator into separated pieces and sum them.


(1) is right. (2) is just wrong because it's wrong. The net electric force between two charged particles acts along a line joining them and its magnitude is given by the coulomb formula. You can take this as an experimental fact. You need to first calculate that force, and then you can calculate its components using trigonometry.

There's no reason to think (2) would work. Examine why you feel it should work. This is a common misunderstanding with students regarding this formula. You have an intuitive sense that if you plug in the x-component of the distance into the coulomb formula, then you'll get the x-component of the force. But this intuitive sense is false. You've seen through the mathematics that the two don't give the same result.

The coulomb formula gives you the magnitude of the net force... the line joining the particles gives the direction. So you have the net force in vector form. When you have the net force you can resolve it into its components. The coulomb formula does NOT give components of the net force.

Now if the r was in the numerator in the coulomb formula and it was a linear term (not squared), then what you're describing would happen to work. But it would just be a mathematical coincidence.


As far as I understand, you attempt to substitute one physical setup (where three charges are present - $A,B,C$) by another one (where there are only two charges are present - $C,D$, where D is the effective charge sitting in the origin). However, in order to do that, one should not forget that this "effective" charge is not equal to $Q$ or $Q\sqrt{2}$. Otherwise the net force will simply diverge in the origin, which does not happen, as we know from the original problem setting. In fact, we have

$F_{net} = 2k\frac{Q^2}{r^2}\cos\theta = k\frac{QQ_{eff}}{(r\cos\theta)^2}$,

from where

$Q_{eff} = 2Q\cos^3\theta$.

This shows that the magnitude of this charge depends on the position of the charge $C$ (quite strange physically, but ok mathematically). As a result, when $\theta\rightarrow \frac{\pi}{2}$, $Q_{eff} \rightarrow 0$. So, model seems to work, however there is still a question whether this new setting is in any way more convenient that the original one...

Hope this helps.

  • $\begingroup$ I don't really think that this is what OP is trying to do. I think he is more confused about vectorial notations and how vectors work. $\endgroup$ – Gonenc Feb 14 '16 at 21:11
  • $\begingroup$ @gonenc: then, this question should be probably moved to math.stackexchange.com. $\endgroup$ – Pavel Feb 14 '16 at 21:27

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.