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Two identical blocks resting on a frictionless, horizontal surface are connected by a light spring having a spring constant $k = 100 N/m$ and an unstretched length $L_i=0.400 m$. A charge $Q$ is slowly placed on each block, causing the spring to stretch to an equilibrium length $L=0.500 m$. Determine the value of $Q$, modeling the blocks as charged particles.

So I was able to solve the problem using particle in equilibrium as follows $$ F_e = k\Delta x \Rightarrow $$ $$ F_e = \frac{k_e Q^2}{L^2} = k(L-L_i) \Leftrightarrow $$ $$ Q= L \sqrt{\frac{k(L-L_i)}{k_e}} = \boxed{ 1.67 \times 10^{-5}C} $$

However, I am trying to solve the problem using conservation of energy, but I am getting a different answer probably due to something wrong I did:

Since the blocks are charged slowly, I ignored kinetic energy: External work done by the force $F_e$ is equivalent to the spring elastic potential energy,

$$ \int_{L_i}^{L} F_e dx = \frac{1}{2}k \left(L-L_i\right)^2 \Rightarrow $$ $$ \int_{L_i}^{L} \frac{k_e Q^2}{x^2} dx = 0.5 \Leftrightarrow $$ $$ \left.\frac{-k_e Q^2}{x}\right|^{0.5}_{0.4} = 0.5 \Leftrightarrow \boxed{Q= 1.05 \times 10^{-5}C } $$ The answers do not match! I assume there is missing energy.

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  • $\begingroup$ Have you noticed that there should be only one factor of L outside the square root in line three of your derivation (instead of two)? Could that be the issue? $\endgroup$
    – Benjamin
    Commented Mar 28, 2016 at 4:15
  • $\begingroup$ @Benjamin that was a typo. I corrected it. $\endgroup$
    – Nazaf
    Commented Mar 28, 2016 at 5:45
  • $\begingroup$ The charges are thought to be brought into the system from infinity so the limits in the last integral should be $-\infty$ and $L$. You will not get the right answer though since the work done to bring the charges together is not equal to the electric potential energy. You are missing the self energies of the point charges which are not as easily computed. $\endgroup$
    – Statics
    Commented Mar 28, 2016 at 9:50
  • $\begingroup$ @Statics How do I integrate from $-\infty$ to $L$? The integral does not even converge ? $\endgroup$
    – Nazaf
    Commented Mar 30, 2016 at 17:06
  • $\begingroup$ It is the last integral. $\lim_{x->\infty} \frac{1}{x}=0$ $\endgroup$
    – Statics
    Commented Mar 30, 2016 at 20:22

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The charge on one block not only repels the charge on the other block, but it also repels the $dq$ charge that you are bringing to charge the system.

As you are slowly charging the system and slowly building up the charges, the force $F_e$ as you integrate is not a constant since the charge is increasing slowly. You may also need to account for the self-energy to build the charged system from scratch (a charge of 0 to 2Q).

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  • $\begingroup$ I see, but how do I account for these kinds of energy in my equation ? $\endgroup$
    – Nazaf
    Commented Mar 28, 2016 at 5:49

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