# Proof that work is the same whaever the path of approach for two charges

I have a quick question on a figure that outlines a proof.

There is a stationary charge $q_1$ and is kept fixed. $q_2$ is moved to the same position along two different paths at a fixed distance away from $q_1$. The work is the same regardless the path of approach.

Work is defined to be:

$$W = \int \vec{F} \cdot d\vec{s}$$

where the work done is just equal and opposite to the Coulomb force. The text accompanying the figure states that $\vec{F}$ has the same magnitude at both paths and is directed radially from $q_1$.

Thus,

$$ds = \frac{dr}{\cos{\theta}} \implies \vec{F} \cdot d\vec{s} = F\cdot dr$$

My question is the line above and how that shows that the work is the same. Would it be related to the dot product of $\vec{F}$ and the differential $d\vec{s}$?

$$\vec{F} \cdot d\vec{s} = Fds\cos{\theta} = F\frac{dr}{\cos{\theta}} \cdot \cos{\theta} = F \cdot dr$$

Oh, since the $\cos{\theta}$ terms cancel out, the angle of approach does not matter since the dot product makes it invariant?

## 2 Answers

First of all, the substitution is not entirely correct, because $ds$ is not the same as $d\vec{s}$, for $ds$ is the modulus of $d\vec{s}$ and is hence a scalar. How to approach the problem is to convert the problem from one of vectors to one of scalars, and thus be able to manipulate the moduli of vectors. Let $v$ be the modulus of $\vec{v}$ to simplify notation. Then,

$W=∫\vec{F}⋅d\vec{s} = ∫F(ds)cos(\theta)$

Where now $\theta$ along with $F$ are functions of $s$. Using the above substitution, that $ds = \frac{dr}{cosθ}$, the integral simplifies to

$W=∫Fdr$

Hence, from $r$ to $r+dr$, the work done through any path is

$W=\int_r^{r+dr}Fdr$

which is constant because $F$ is a function of r. Infact, you can use coloumb's law and plug in F to see what the work value is.

Update: Derivation of substitution Again considering only the moduli of vectors, consider a vector $\vec{s}$, which makes an angle $\theta$ with the radial axis (line connecting head of $\vec{s}$ with center of force). Then, the magnitude of the radial component of the vector, from vector decomposition, is $r = s*cos(\theta)$. Hence, $r/s = cos\theta$, and taking the limit as $s$ goes to zero, we get the differential form, $dr/da = cos\theta$ or $ds = \frac{dr}{cos\theta}$

• Thanks for your answer. Could you explain a bit how they derived the substitution for $ds$ as well? And since the force becomes a constant, we can conclude that it is the same for any angle at a distance $r + dr$?
– user43617
May 19, 2015 at 0:50
• Answered both parts. The force is not constant, but that is irrelevant. See the addition I have put. May 19, 2015 at 1:01

Since this is not a homework, I want to add some thoughts about the limits for this calculations.

First, every body, if accelerated, radiates and this happens on every curved path (if the path is not the geodesic gravitational line).

Second, every charged particle has a magnetic dipole moment too and this moments get aligned under the each other influence. This is a loss of energy too.

It is a little bit sophisticated. But it is true.

• Could you provide some references for this?
– user43617
May 19, 2015 at 10:46
• @Jun-Goo Kwak Not the same but related to magnetic dipole moments: physics.stackexchange.com/questions/185121/… May 21, 2015 at 11:22