- #1

Karol

- 1,380

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## 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 q

_{2}. what is it's velocity 1[cm] from q

_{1}.

## 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?