Why, in this solution, acceleration is constant even when it depends on distance between two charges? I used integration of $a=dv/dt$ to solve this Why, in this solution is acceleration constant, even when it depends on the distance between two charges? I used integration of $a=dv/dt$ to solve this.
Question
Two particles have equal masses of $5.0 \ g$ each and opposite charges of $+4.0 \times 10^{-5} C$ and $-4.0 \times 10^{-5} C$. They are released from rest with a separation of $1.0 \ m$ between them. Find the speeds of the particles when the separation is reduced to $50 \ cm$.
This involves Coulomb's law, Newton's 2nd law of motion and kinematics of relative acceleration.
Solution of above question
$$q_1 = q_2 = 4 \times 10^{-5}C \ \ \ and \ \ \ s=1m, \ \ m=5g=0.005 kg$$
$$F=K \frac{q^2}{r^2} = \frac{9 \times 10^9 \times (4 \times 10^{-5})^2}{1^2} = 14.4 \ N$$
$$Acceleration \ \ a = \frac{F}{m} = \frac{14.4}{0.005}=2880 \ m/s^2$$
$$Now \ \ u = 0 \ \ \ \ s = 50 \ cm = 0.5 \ m, \ \ \ a = 2880 \ m/s^2, \ \ \ v = \ ?$$
$$v^2 = u^2 + 2as \ \ \ \rightarrow \ \ v^2 = 0 + 2 \times 2880 \times 0.5$$
$$v = \sqrt{2880} = 53.66 \ m/s \approx 54 \ m/s \ \ \ for \ each \ particle.$$
 A: Actually, acceleration is not constant in this case because in time $dt$ the force would change. So, the acceleration also changes even in time $dt$. I think the solution is wrong but the answer is correct.
If you go for energy conservation, which doesn't depend on acceleration you get the same answer:
$$ K_{1} + U_{1} = K_{2} + U_{2}$$
$K_{1} = 0$, since they are at rest initially, where:
$$U_{1} = - \frac{k q^{2}}{r}$$
$K_{2} = m v^{2}$ (you need to see this carefully, it is $m v^{2}$ not $\frac{1}{2} m v^{2}$, because you have to take the KE (kinetic energy) of the system, since two particles are there each moving with $\frac{1}{2} m v^{2}$. So:
$$2 \frac{1}{2} m v^{2} = m v^{2}$$
They move with different KE if their mass is different, so then you need to use momentum conservation to find the relation between their speed and proceed:
$$U_{2} = - \frac{2k q^{2}}{r}$$
So, if you substitute these in that you will give the answer to be $24 \cdot \sqrt{5}$, which gives the same answer.
A: The method used in the given solution is completely incorrect.  Its only redeeming virtue is that it happens to give the correct answer through a numerical coincidence.  The simplest way to solve the answer correctly would involve energy conservation, as described in Visza Sekar's answer.
To be a bit more concrete about how this happened, let's generalize to a situation where initial distance of the charges is $r_i$ and the final distance is $r_f$.  Following the method in the given solution, we would find that
$$
mv^2 = 2 \frac{kq^2}{r_i^2} (r_i - r_f)
$$
while the (correct) energy conservation method would yield
$$
mv^2 = kq^2 \left( \frac{1}{r_f} - \frac{1}{r_i} \right).
$$
A bit of algebra shows that these two expressions are equal only when $r_f = r_i/2$, which happens to be the case in this problem.  However, if the problem involved an initial and final separation that were not related like this, then the two methods would give different result, and the result from the solution's method would be wrong.
A: The total change in field energy equals the negative of the  total amount of work done on all charges.
For 2 point charges, the total change in field energy is just the change in potential energy between them
$$(U_{2}-U_{1}) = -(\Delta K_{q_{1}} + \Delta K_{q_{2}})$$
$$U_{2} - U_{1} = \frac{1}{4\pi\epsilon_0}\frac{ q_{1} q_{2}}{0.5} -\frac{1}{4\pi\epsilon_0}\frac{ q_{1} q_{2}}{1}
  $$
Now because we know the masses are equal, and the situation is symmetric. We know that the change in kinetic energy of both charges is the same, as we say that they are both released at the same time.
$$ \Delta K_{q_{1}} = \Delta K_{q_{2}}$$
Which allows us to write.
$$(U_{2}-U_{1}) = -(2\Delta K_{q})$$
Meaning:
$$-\frac{U_{2}-U_{1}}{2} = \Delta K_{q}$$
$$-\frac{U_{2}-U_{1}}{2} = \frac{1}{2}mv^2$$
$$\sqrt{-\frac{U_{2}-U_{1}}{m}} = v$$
Notice if we set the change in kinetic energy of one of them to be zero, this is is same as fixing one in place, which can be intutatively thought of just the change of potential equals the negative of the total amount of work done on a particle, which is commonly taught. This relies on electrostatics where the other charge is fixed.
More handwavy, you can say that the potential energy is mutually shared so the change in potential of any one charge is halved
Haven't plugged in numbers but if the comment above is correct, it yields then same answer, but have no idea why, it shouldnt, probably is the same as the average acceleration halfed.
