Question

Two parallel plates are a distance apart with a potential difference between them. a point charge moves from the negatively charged plate to the positively charged plate. The charch gains kinetic energy w. The distance between the plates is doubled and the potential difference between them is halved what is the kinetic energy gained by an identical charge moving between those plates?

Answer 1

Answer:

We can relate the kinetic energy of the particle to the potential difference between the plates by following equations:

Work energy theorem:

$W_{total} = \Delta K = K_2 - K_1 = w - 0$

$W = Fx\\F = W/d = w/d$

$F = Eq \\E = F/q = w/(dq)$

$V = Ed = \frac{w}{dq} d = w/q$

So,

$w = Vq$

If the distance is doubled and the potential difference is halved, then

$w = qV/2$

Explanation:

As can be seen from the relationship between kinetic energy and the potential difference, the distance between the plates has no effect on the relation between kinetic energy and the potential difference. Since the charge of the second particle is equal to that of the first one, the new kinetic energy would be half of the first kinetic energy.