Electron moving between 2 point charges

[Karol]

Homework Statement

A point charge $$q_1=+20\cdot 10^{-9}[Coulomb]$$ is 5[cm] distance from charge $$q_2=-12\cdot 10^{-9}[Coulomb]$$.
An electron is released from 1[cm] distance from q₂. what is it's velocity 1[cm] from q₁.

Homework Equations

The potential=Voltage from a point charge: $$V=K\frac{q}{r}$$
The constant $$K=9\cdot 10^9$$
The work done to move from one point in the field to another: $$W=V\cdot q$$
The electron charge: $$e=1.6\cdot 10^{-19}[Coulomb]$$.
The electron mass: $$m_e=9.11\cdot 10^{-}[Kg]$$

The Attempt at a Solution

I solved:
I took the signs of the charges and calculated the potential:
$V_1=9\cdot 10^9\left(\frac{-12\cdot 10^{-9}}{0.01}+\frac{20\cdot 10^{-9}}{0.04}\right)=9\left(\frac{20}{0.04}-\frac{12}{0.01}\right)$
$V_2=9\left(\frac{20}{0.01}-\frac{12}{0.04}\right)$
Then i subtracted the potentials and translated the work into kinetic energy and found the velocity. it was wrong. but if i change the signs of the charges it comes out like in the book:
V=8.7E7[m/sec].
The solution that works:
$V_1=9\left(\frac{-20}{0.04}+\frac{12}{0.01}\right)$
$V_2=9\left(\frac{-20}{0.01}+\frac{12}{0.04}\right)$
Why does it work?

 

[Doc Al]

Did you take into account the sign of the electron charge?

 

[Karol]

No, why should i? for the potential it doesn't matter, in the equation $$V=K\frac{q}{r}$$ only the stationary charge appears.
The only difference the electron charge makes is whether i invest work or get work, that's all, no? the number i will get for the amount of work will be the same

 

[ehild]

You need the potential energy of the electron, when calculating its final kinetic energy. The potential energy is charge times potential, you need to multiply the potential difference with the charge of the electron, 1.6x10^-19 C.If you do not take the sign of the electron charge into account, you get a negative value for the kinetic energy.
It is an other thing that the velocity of the electron too high, you should apply SR.

 

[Karol]

I know that $W=V\cdot q$, that the work done is the potential times the charge and i know that i search for the potential energy of the electron and i calculated the velocity, i just didn't write all that because the problem was that i got a different value, not sign, for the kinetic energy when i changed the signs in the formula, just to test.
The potential at a point doesn't depend on the charge that is placed later on that point, so, when calculating the potential V at the start and end points of the electron's path i don't need, yet, to take into account nor it's sign and nor it's charge.

 

[ehild]

If you calculate the kinetic energy of the electron from the absolute value of the potential difference multiplied by the elementary charge, the sign of the electron charge does not matter.
Better to show the details of your calculation. You can have a sign error when determining the potential difference.

 

[Karol]

The electron moves from A to B.
$V_A=9\cdot 10^9\left(\frac{-12\cdot 10^{-9}}{0.01}+\frac{20\cdot 10^{-9}}{0.04}\right)=9\left(\frac{20}{0.04}-\frac{12}{0.01}\right)=-6300[V]$
$V_B=9\left(\frac{20}{0.01}-\frac{12}{0.04}\right)=15,300[V]$
$\Delta V=9000[V]$
$W=\Delta V\cdot e=E_k=9000\cdot 1.6\cdot 10^{-19}=1.44\cdot 10^{-15}=\frac{1}{2}\cdot 9.11\cdot 10^{-31}\cdot V^2\rightarrow V=5.62\cdot 10^7[\frac{m}{s}]$
It should be 8.7E7[m/s]

 

[ehild]

The electron moves from A to B.
$V_A=9\cdot 10^9\left(\frac{-12\cdot 10^{-9}}{0.01}+\frac{20\cdot 10^{-9}}{0.04}\right)=9\left(\frac{20}{0.04}-\frac{12}{0.01}\right)=-6300[V]$
$V_B=9\left(\frac{20}{0.01}-\frac{12}{0.04}\right)=15,300[V]$
$\Delta V=9000[V]$

You got the mistake here.
$\Delta V= V_B-V_A=15300-(-6300)=216000[V]$

 

[Karol]

Thankks!