Question

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a distance of 46.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

I have no clue what do to...I am at a complete loss!

Answer 1

`7.8\cdot 10^6 m/s`

Step-by-step explanation:

When the electron moves, the gain in its kinetic energy is equal to the decrease in its electric potential energy, since the total energy must be conserved.

Therefore, we can write:

`K_f-K_i = U_i-U_f`

where

`K_f=(1)/(2)mv^2` is the final kinetic energy of the electron, where

`m=9.11\cdot 10^(-31) kg` is its mass

v is its final speed

`K_i=0` is the initial kinetic energy (the electron starts from rest)

`U_i=(kq_1 e)/(r_1)+(kq_2 e)/(r_2)` is the initial electric potential energy, where

k is the Coulomb's constant

`q_1=3.90\cdot 10^(-9)C` is the charge 1

`e=-1.6\cdot 10^(-19)C` is the electron charge

`r_1=0.23 m` is the initial distance of the electron from charge 1

`q_2=1.80\cdot 10^(-9)C` is charge 2

`r_2=0.23 m` is the initial distance of the electron from charge 2

`U_f=(kq_2 e)/(r'_1)+(kq_2 e)/(r'_2)` is the final electric potential energy, where

`r_1=0.10 m` is the final distance between electron and charge 1

`r_2=0.46-0.10 = 0.36 m` is the final distance between electron and charge 2

Substituting everything we find:

`U_i = ((9\cdot 10^9)(3.90\cdot 10^(-9))(-1.6\cdot 10^(-19)))/(0.23)+((9\cdot 10^9)(1.80\cdot 10^(-9))(-1.6\cdot 10^(-19)))/(0.23)=-3.57\cdot 10^(-17) J`

`U_f = ((9\cdot 10^9)(3.90\cdot 10^(-9))(-1.6\cdot 10^(-19)))/(0.10)+((9\cdot 10^9)(1.80\cdot 10^(-9))(-1.6\cdot 10^(-19)))/(0.36)=-6.34\cdot 10^(-17) J`

So the final kinetic energy is

`K_f=U_i-U_f=-3.57\cdot 10^(-17)-(-6.34\cdot 10^(-17))=2.77\cdot 10^(-17) J`

And therefore, the final speed is:

`v=\sqrt{(2K_f)/(m)}=\sqrt{(2(2.77\cdot 10^(-17)))/(9.11\cdot 10^(-31))}=7.8\cdot 10^6 m/s`