Work done in assembling charges

Question

We have this system here

Says that the total energy required to assemble the system is \begin{align} \text{0 for $q_1$}\end{align}

\begin{align} \frac{kq_1 q_2}{l} &&\text{for $q_2$} \tag 1\\ \frac{kq_1q_3}{l}+\frac{kq_2q_3}{l} && \text{for $q_3$} \tag 2\\ \end{align}

Therefore the total work done is

But why is the work done on $q_3$ equal to \begin{align} \frac{kq_1q_3}{l}+\frac{kq_2q_3}{l} \end{align} It should be product of net force and net displacement, right?
(It should take the resultant of forces)

But instead it adds the work done against each individual force which seems wrong to me.

After all if the horizontal components are cancelling each other, it reduces the opposing field which should just reduce work done by us. So where is the extra work used?

Similarly in this situation

But it should be $$U_i = \frac{(kqQ_2)}{2R}$$

$$U_f = \frac{(kqQ_1)}{2R}$$

As calculated by integrating for electric field

So basically my question is Why should the work of inividual forces be considered instead of the net force in assembling a system of charges?

Edit:
Just a suggestion
Will this integral work: $$\int_{\infty}^{\frac{\sqrt3l}{2}}(\vec F_1+\vec F_2) .dx $$ If no can an integral be defined using net force

Answer 1

Assembling a system means taking the different components and transporting them from a point infinatly far away to the desired point in the system.

The first particle will have no other particles to interact with and therefore the cost of placing it in the system will be zero.

The second particle will interact with the first particle according to the Coulomb force. To obtain the work required to place the second particle you have to integrate the Coulomb force along the path (from $\infty$ to l).

$$ \int^l_\infty -k\frac{q_1q_2}{r^2}dr = -kq_1q_2\left[-\frac{1}{r}\right]^l_\infty = k\frac{q_1q_2}{l} $$

The third particle has to interact with both particle 1 and 2. We now have to integrate over both forces:

$$ \int^l_\infty -k\frac{q_1q_3}{r^2}dr + \int^l_\infty-k\frac{q_2q_3}{r^2}dr = -k(q_1q_3 + q_2q_3)\left[-\frac{1}{r}\right]^l_\infty = k\frac{q_1q_3}{l} + k\frac{q_2q_3}{l} $$

The sum of all the work required to assemble the system will be:

$$ k\frac{q_1q_2}{l} + k\frac{q_1q_3}{l} + k\frac{q_2q_3}{l} $$

Answer 2

Building up the distribution:

The work required to bring $Q_{2}$ in the presence of $Q_{1}$

$ \frac{Q_{2}Q_{1}}{4\pi\epsilon_{0}R_{1,2}}$

Where $R_{1,2}$ is the distance between charges 1 and 2

The work required to bring $Q_{3}$ in the presence of fields generated by $Q_{2}$ and $Q_{1}$ from infinity to its location, would be:

$\int -(\vec{F_{1}} + \vec{F_{2}}) \cdot \vec{dl}$

$\int -\vec{F_{1}} \cdot \vec{dl_{1}} + -\int\vec{F_{2}} \cdot \vec{dl_{2}}$

$\int -\frac{Q_{3}Q_{1}}{4\pi\epsilon_{0}r_{1}^2}\hat r_{1} \cdot \vec{dl_{1}} + \int -\frac{Q_{3}Q_{2}}{4\pi\epsilon_{0}r_{2}^2}\hat r_{2} \cdot \vec{dl_{2}}$

Note the path $\vec{l_{2}}$ and $\vec{l_{2}}$ are the same path, but are both defined in terms of the radial distance from each charge since we are working with 2 different variables$(r_{1},r_{2})$, so we need to define 2 different parametizations in order to integrate.

Since we are evaluating the field for each charge as a function of radial distance, the line element in spherical coordinates would be:

$\vec{dl_{1}} = dr_{1} \hat r_{1} + r_{1} d\theta_{1} \hat \theta_{1} + r_{1} sin(\theta_{1}) d\phi_{1}\hat \phi_{1}$

$\vec{dl_{2}} = dr_{2} \hat r_{2} + r_{2} d\theta_{2} \hat \theta_{2} + r_{2} sin(\theta_{2}) d\phi_{2}\hat \phi_{2}$

Since the force from each charge only has a $\hat r_{1.2}$ component, the dot product of each path with its respective force leaves only the the first component of each path. ( as the rest dissapear)

Which gives us

$\int -\frac{Q_{3}Q_{1}}{4\pi\epsilon_{0}r_{1}^2} dr_{1} + \int -\frac{Q_{3}Q_{2}}{4\pi\epsilon_{0}r_{2}^2} dr_{2}$

What are the bounds for each respective integral? Well the path that we want is a path from infinity to the location of the charge. The variables as we have it, are in terms of radial distance from each charge( as the other components vanish in the dot product) so the bounds are:

$r_{1}:$ $\infty$ to $R_{1}$, where $R_{1}$ is the final radial distance from $Q_{1}$

$r_{2}:$ $\infty$ to $R_{2}$, where $R_{2}$ is the final radial distance from $Q_{2}$

$\int_{\infty}^{R_{1}} -\frac{Q_{3}Q_{1}}{4\pi\epsilon_{0}r_{1}^2} dr_{1} + \int_{\infty}^{R_{2}} -\frac{Q_{3}Q_{2}}{4\pi\epsilon_{0}r_{2}^2} dr_{2}$

$ \frac{Q_{3}Q_{1}}{4\pi\epsilon_{0}R_{1}} +\frac{Q_{3}Q_{2}}{4\pi\epsilon_{0}R_{2}}$

Adding the total amount of work gets us

$\frac{Q_{2}Q_{1}}{4\pi\epsilon_{0}R_{1,2}} + \frac{Q_{3}Q_{1}}{4\pi\epsilon_{0}R_{1}} +\frac{Q_{3}Q_{2}}{4\pi\epsilon_{0}R_{2}}$