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We have this system here enter image description 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 isThis is for scalene triangle

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

enter image description here

enter image description here

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

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  • $\begingroup$ Sorry about the text $\endgroup$
    – Aurelius
    Commented Mar 29, 2022 at 6:14
  • $\begingroup$ Could not find how to add subscripts and superscripts $\endgroup$
    – Aurelius
    Commented Mar 29, 2022 at 6:14
  • $\begingroup$ Try to use latex for embedded formulas. $\endgroup$ Commented Mar 29, 2022 at 6:17
  • $\begingroup$ Also if the work done is same either way, could anyone provide a proof that integrating either way leads to same results $\endgroup$
    – Aurelius
    Commented Aug 25, 2023 at 21:23

2 Answers 2

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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

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

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  • $\begingroup$ Can you tell me if the integral I suggested later is correct? ( It's kind of like finding a centre of charge and using the net field of both charges) $\endgroup$
    – Aurelius
    Commented Aug 28, 2023 at 18:14
  • $\begingroup$ I am unsure what "a" is in your formula, however I assume as you've stated this is the position we can replace the distribution with a single charge. This is incorrect as this would find the potential due to the field at the location of the "center of charge", not the total energy of the system. I also am unaware of the quantity "center of charge" having any validity when determining the energy of a distribution. How have you come to this method? $\endgroup$ Commented Aug 28, 2023 at 22:48
  • $\begingroup$ As a single integral, if we only focus on continuous distribution instead of point charges, the energy in the EM field is $$\iiint_{V} \frac{1}{2}\epsilon_{0} |\vec{E}|^2 dv$$ $\endgroup$ Commented Aug 28, 2023 at 22:55
  • $\begingroup$ I came to that integral by :, I just found the net force on the third charge by the two charges and assumed that force was exerted by a point charge of (whatever magnitude gives the same value of force) placed at the midpoint of the base. 😅 $\endgroup$
    – Aurelius
    Commented Aug 29, 2023 at 6:05
  • $\begingroup$ This wouldn't make sense , you aren't taking into the account of the placing of the other charges, the distribution can not be replaced by a single charge,this approximation only holds at large distances, and finally we aren't placing the third charge at this center location $\endgroup$ Commented Aug 29, 2023 at 11:07
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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:

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

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  • $\begingroup$ Remember that $\frac{1}{r^2}$ is the radial distance from the individual charge, so it cannot be factored as they are 2 seperate variables. ( 1 for each charge). Its best to seperate the 2 integrals and work out the path for each variable and then add them up. (Or doing it your way find out one variable in terms of the other) $\endgroup$ Commented Mar 29, 2022 at 11:54
  • $\begingroup$ You're right, but only the radial part of the path matters, since no work is created on a perpendicular force, so I thought it would be overcomplicating the problem. But as you suggested I have split the integral into two seperate. $\endgroup$ Commented Mar 29, 2022 at 13:38
  • $\begingroup$ I'd edit the variables and change label $l_{1}, l_{2}$ $\endgroup$ Commented Mar 29, 2022 at 14:00
  • $\begingroup$ Another question: physics.stackexchange.com/questions/381273/…. As shown by the answers to this question, should the work done not be double of your answer because there will be some heat loss of some kind and to counter the repulsion on the already assembled charge? $\endgroup$
    – Aurelius
    Commented May 17, 2023 at 15:16
  • $\begingroup$ @MartinGardfjell 👆 $\endgroup$
    – Aurelius
    Commented May 26, 2023 at 20:07

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