- #1
Heisenberg7
- 101
- 18
- Homework Statement
- Two identical point charges are moving in free space, when they are 60 cm apart, their velocities are equal in magnitude and make angles of 45 degrees from the line joining them as shown in the figure. If at this instant their total kinetic energy is equal to their potential energy, what will be the distance of closest approach between them? (Source: YouTube video; Name: Electrostatics | Advanced Problem | Closest Distance of Approach Between Two Charged particles)
- Relevant Equations
- $$v_f = v_i - at$$
$$s = v_ot - \frac {at^2} 2$$
$$F = ma$$
A few hours ago, I tried solving this problem and I'm still not quite sure if I've made a mistake somewhere or perhaps the guy in the video is wrong? Anyway, here's my solution:
In the problem we're given that the total kinetic energy of of these 2 charges at this instant(look at the picture) is equal to their potential energy, thus we can write:$$\sum E_k=Ep$$ $$2 \frac {mv_o^2} 2 =k \frac {q^2} r$$ Simplifying this we get $$mv_o^2 = k \frac {q^2} r , (1)$$
The speed of one particle in the y direction is going to stay constant during the whole interaction because the force on the particles does work only in the x direction. So we can write $$v_y=v_{oy}=v_o \cos 45^{\circ}=v_o \frac {\sqrt{2}}2$$ Now the x direction. The initial velocity in the x direction is equal to the initial velocity in the y direction(magnitude). Both of these particles are going to decelerate due to the electrostatic force acting on both particles $$F_e=k \frac {q^2} {r^2}$$ By the second Newton's law, we get that the acceleration is, $$a=k\frac {q^2} {mr^2}$$ When the magnitude of velocities of these 2 particles reaches zero, we get the closest distance of approach. By plugging that into the kinematics equation we get, ##v_i=at##, or $$v_{ox}=at \iff v_o \frac {\sqrt{2}} 2=at \iff v_o \frac {\sqrt{2}} 2=k\frac {q^2} {mr^2} t$$ $$\iff t = \frac {\sqrt{2}} 2 \frac{v_o m r^2} {kq^2}$$Using the second kinematics equation we will get the distance covered by one of the particles, $$s=v_{ox}t−\frac {at^2} 2$$, or $$s=\frac{\sqrt{2}} 2 v_o t−\frac {at^2} 2$$ Plugging in for ##t## and ##a## and simplifying we get $$s=\frac{1} 2 \frac {v_o^2mr^2}{kq^2}−\frac {1} 4 \frac {v_o^2mr^2} {kq^2}$$ From this we get $$s=\frac {1} 4 \frac {mv_o^2 r^2}{kq^2}$$ Now comes the equation we derived above (1). By substituting, we get $$s=\frac {1} 4 \frac {k \frac {q^2} {r} r^2} {kq^2}$$ Or $$s=\frac{1} 4 r$$ Both particles are going to travel the same distance thus the closest distance of approach is going to be $$r_2=r−2s \iff r_2=r−2\frac {1} 4 r \iff r_2=r−\frac {1} 2 r \iff r_2=\frac {1} 2 r \iff r_ 2=\frac {1} 2 60cm \iff r_2=30cm$$ In the video the guy somehow got $$r_2=\frac {2} 3 r$$
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