Work done by electric field, $W$ is the negative of the change in electric potential energy, $U_r$. $$W=-\Delta U$$ By considering an infinitesimal change in EPE, we can deduce that electric force, $F_e$ is the negative potential energy gradient. $$-dU_r=F_e\cdot dr$$ $$-\frac{dU_r}{dr}=F_e$$ where $$U_r=\frac{kQq}{r}$$ But from the first equation above, we can say that an infinitesimal change in EPE is the same as infinitesimal change in work. $$dW=F_e\cdot dr$$ $$\frac{dW}{dr}=F_e$$ This contradicts with the first derivation as $W$ and $U_r$ both have the same equation, and the negative in the derivative is missing. Work done by electric field to move a charge from r to $\infty$ : $$W=-\Delta U_r=\int_{r}^{\infty}\frac{kQq}{r^2}dr$$ $$-(U_\infty - U_r)=\frac{kQq}{r}$$ $$U_r=\frac{kQq}{r}$$ $$W=U_r$$ Finding $F_e$, $$-\frac{dU_r}{dr}=-\frac{d}{dr}(\frac{kQq}{r})=\frac{kQq}{r^2}$$ $$\frac{dW}{dr}=\frac{d}{dr}(\frac{kQq}{r})=-\frac{kQq}{r^2}$$
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4$\begingroup$ From $-dU_r / dr = F_e$ and $dW/dr = F_e$ it follows $-dU_r/dr = dW/dr$ which equals $dW = -dU_r$. This is exactly what you wrote in the first line $W = -\Delta U$. What do you mean by "this contradicts with the first derivation as $W$ and $U_r$ both have the same equation"? $\endgroup$– Marko GulinCommented Apr 1, 2022 at 12:34
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$\begingroup$ $W$ and $U_r$ equals $\frac{kQq}{r}$, and by differentiating this, we get different equations of $F_e$, as the sign is different which is mathematically obvious as there is a missing negative sign in the derivative of work done. $\endgroup$– radastroCommented Apr 1, 2022 at 12:45
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$\begingroup$ This might be a possible reason: $W=-\Delta U_r$, only $U_r$ can be express as a change, and we can then relate this to the derivative, whereas $W$ already means a change, so we don't write $dW$ $\endgroup$– radastroCommented Apr 1, 2022 at 12:56
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$\begingroup$ Am I correct to assume that you are confused how come $W$ and $dW$ mean the same thing? If yes, then see the post below. $\endgroup$– Marko GulinCommented Apr 1, 2022 at 13:13
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$\begingroup$ Who says that $W$ is the Coulomb potential? Anyways: When working with work, always make sure to inspect who performs work on whom before discussing the sign! $\endgroup$– kricheliCommented Apr 1, 2022 at 13:18
1 Answer
The mathematics in your post is correct:
$$\text{from} \qquad F_e = -\frac{d U_r}{dr} \qquad \text{and} \qquad F_e = \frac{dW}{dr} \qquad \text{it follows} \qquad \boxed{dW = -dU_r} \tag 1$$
As you already mentioned, the work $W$ done by a conservative force equals the negative of the change in potential energy $\Delta U$
$$W = -\Delta U \tag 2$$
What is seems to me is that you are confused how come $dW$ in Eq. (1) and $W$ in Eq. (2) mean the same thing. Note that the well known equation for work
$$W = \int \vec{F} \cdot d\vec{r} \tag 3$$
actually comes from
$$dW = \vec{F} \cdot d\vec{r} \tag 4$$
You can read the Eq. (4) as follows: a force $\vec{F}$ over infinitesimal small displacement $d\vec{r}$ does infinitesimal small work $dW$.
Responding to your edit, you are getting the wrong sign because independent variable $r$ in
Work done by electric field to move a charge from r to $\infty$ : $$W=-\Delta U_r=\int_{r}^{\infty}\frac{kQq}{r^2}dr$$
is lower bound of the integral. Remember that by definition, if function $f$ is continuous on interval $[a,x]$ then
$$F(x) = \int_{a}^{x} f(t) dt \quad \rightarrow \quad F'(x) = f(x)$$
But in your case you have
$$F(x) = \int_{x}^{a} f(t) dt = - \int_{a}^{x} f(t) dt \quad \rightarrow \quad -F'(x) = f(x)$$
The expression for work done by electric field to move a charge from $r$ to $\infty$ that you found
$$U_r = \frac{k Q q}{r}$$ $$W = U_r$$
is correct. But when you take derivate of that expression to find force $F_e$, you have to take into account that $r$ was actually lower bound.
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1$\begingroup$ $W_{\rm final}-W_{\rm initial}$ has no meaning in Physics. $\endgroup$– FarcherCommented Apr 1, 2022 at 13:31
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$\begingroup$ @Farcher You are right, better not to use (final) and (initial) in the context of work. $\endgroup$ Commented Apr 1, 2022 at 13:52