Problema 72-3    Source:  Problem 29.3 - page 905 - KNIGHT, Randall D. -    Book: Física - Uma Abordagem Estratégica - 2ª Edição - Ed. Bookman - 2009.

A proton departs from the positive plate's rest in a parallel plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the final speed of the proton if the experiment is repeated with twice the amount of original charge on each capacitor plate.

Solution of the Problem 72-3   

It is important to note that when we double the charge of a capacitor, the energy accumulated by it is multiplied by 4, as we can infer by eq. 03-07, repeated below for further understanding. Naturally, we are considering that there has been no change in its capacitance.

Considering the energy that the proton acquired in the first experiment as W₁, so for the second experiment we have W₂ = 4 W₁. Evidently, energy W₁ is equal to the variation of kinetic energy to which the proton was subjected. We can write this as

So for the second experiment we can write

Dividing  eq. 72-03.2   by  eq. 72-03.1   we get

Arranging the terms, we have

How do we know that W₂ = 4 W₁, so we can easily conclude that

Attention -   In the book there is a typo in the answers section, which contains the value √2 v₁. Or they forgot to take into account twice the charge on the capacitor when solving the problem.