Return to 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 $dW$.

As for your edit, you are getting the wrong sign because independent variable $r$ is the 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

$$F(x) = \int_{x}^{a} f(t) dt = - \int_{a}^{x} f(t) dt \qquad \rightarrow \qquad F'(x) = -f(x)$$

You just missed the negative sign when you wrote "$\frac{dW}{dr}$" under "Finding $F_e$".

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.

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 the 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 $dW$.

As for your edit, you are getting the wrong sign because independent variable $r$ is the lower bound. Remember that by definition

$$F(x) = \int_{a}^{x} f(t) dt \qquad \rightarrow \qquad F'(x) = f(x)$$

But in your case you have

$$F(x) = \int_{x}^{a} f(t) dt = - \int_{a}^{x} f(t) dt \qquad \rightarrow \qquad F'(x) = -f(x)$$

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 $dW$.

As for your edit, you are getting the wrong sign because independent variable $r$ is the 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

$$F(x) = \int_{x}^{a} f(t) dt = - \int_{a}^{x} f(t) dt \qquad \rightarrow \qquad F'(x) = -f(x)$$

You just missed the negative sign when you wrote "$\frac{dW}{dr}$" under "Finding $F_e$".

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 the work

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

is a shorter form of

$$W_f - W_i = \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r} \tag 3$$

This 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 $dW$. If you integrate the above equation

$$\int_{W_i}^{W_f} dW = \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r}$$

you should get the result in Eq. (3).

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 the 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 $dW$.

As for your edit, you are getting the wrong sign because independent variable $r$ is the lower bound. Remember that by definition

$$F(x) = \int_{a}^{x} f(t) dt \qquad \rightarrow \qquad F'(x) = f(x)$$

But in your case you have

$$F(x) = \int_{x}^{a} f(t) dt = - \int_{a}^{x} f(t) dt \qquad \rightarrow \qquad F'(x) = -f(x)$$