Why do we need to take the absolute value of charge in an electron gun?

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For part (a) of this problem,

The solution is,

However, why did they need to take the absolute value of the charge? I thought they could keep the original signs as shown below:

Using energy conservation, the electric force dose internal work transferring electric potential energy into kinetic energy

Thanks for any help!

 

[haruspex]

The velocity before the electron transited the potential gradient was zero. Your energy equation should read $\Delta KE+\Delta EPE=0$.
To accelerate the electrons, the gun must have a negative potential, so $\Delta V=+2500V$.
This gives $\Delta KE=-q\Delta V$.
The given solution used the knowledge that the field increases the KE in a different way. Instead of taking the trouble to manage the signs correctly, it just forced everything positive.

 

[ChiralSuperfields]

To accelerate the electrons, the gun must have a negative potential, so $\Delta V=+2500V$.

Thank you @haruspex! Are you saying that V_i = -2500V and V_f = 0? Why must gun have negative potential and therefore Delta V = 2500V?

 

[haruspex]

Why must gun have negative potential

So as to repel the electrons.

 

[ChiralSuperfields]

So as to repel the electrons.

Thanks for the reply @haruspex! I'm assuming that at V_i = -2500V there is no protons and a finite number of electrons. And at V_f = 0 there is no electrons and protons, correct?

However, if the gun has negative potential, then the electric potential energy of the electrons will increase over the gun as delta V is positive. However, how is this possible since the kinetic energy also increases? Where dose this energy that turns into kinetic energy come from?

 

[haruspex]

if the gun has negative potential, then the electric potential energy of the electrons will increase over the gun as delta V is positive

Don’t confuse electric potential with electric potential energy.
When a negative charge moves to a higher potential it loses electric potential energy. $\Delta E=q\Delta V$.

 

[ChiralSuperfields]

Don’t confuse electric potential with electric potential energy.
When a negative charge moves to a higher potential it loses electric potential energy. $\Delta E=q\Delta V$.

Oh got it now! Thanks again @haruspex!