Colliding charged plates !

Electricity and Magnetism Level 5

Two fixed, identical conducting plates $α$ & $β$ , each of surface area $S$ are charged to $-Q$ and $q$ , respectively, where $Q > q > 0$ . A third identical plate ( $γ$ ), free to move is located on the other side of the plate with charge $q$ at a distance $d$ . The third plate is released and collides with the plate $β$ . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst $β$ & $γ$ .

Find the velocity of the plate $γ$ after the collision and at a distance $d$ from the plate $β$ .

If $v=(Q-\frac{q}{x})(\frac{d}{m \epsilon_0 S})^{\frac{y}{z}}$

then, find $(x+y+z)$ .

The answer is 5.000.

2 solutions

Вук Радовић
Jun 16, 2015

First there is a net force on third plate

$F=\frac{Q(Q-q)}{2\epsilon_0 S}$

Now you can find the speed of plate in the moment of the collision.

$v_1^2=\frac{Q(Q-q)}{2\epsilon_0m}d$

So now there is a key part of the problem during the collision the charge will redistrubate over the second and third plates. Let $q_1$ be the new charge of second and $q_2$ the new charge of the third plate. While they are connected net eletric field between them is 0.

$=> -Q+q_1-q_2=0$

And we know also that:

$q+Q=q_1+q_2$

$=> q_1=Q+q/2$ and $q_2=q/2$

Now if you do the same thing with forces from the beginning you will easy find:

$v_f=(Q-q/2)\sqrt{\frac{d}{2\epsilon_0mS}}$

Yeah did the same nice problem!

Harsh Shrivastava - 3 years, 12 months ago

Ace Pilot
Aug 30, 2015

V²=(Q-q)Qd/A€m Let Q1 and be the charge on the left face of left plate then writing charge on all faces, when plates are in state of collision, it will act as a single plate with charge q+Q and when they seperate their inner surfaces will be without charge Again V²-U²=2F/m.d

Colliding charged plates !

The answer is 5.000.

2 solutions

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