Answer:
v = 4.31*10^6 m/s
Explanation:
In order to calculate the speed of the electron, you first calculate the acceleration of the electron, generated by the electric force.
You use the following formula:
[tex]F=ma=qE[/tex] (1)
m: mass of the electron = 9.1*10-31kg
a: acceleration of the electron = ?
q: charge of the electron = 1.6*10^-19C
E: electric field
Then, it is necessary to calculate the electric field, which is given by:
[tex]E_T=k\frac{q_1}{r_1^2}-\frac{q_2}{r_2^2}[/tex] (2)
The electric field is calculated at the point midway of the two charges q1 and q2.
The minus sing means that the electric force are opposite in such a point.
q1: charge 1 = 4.00nC = 4.00*10^-9 C
q2: charge 2 = 1.55nC = 1.55*10^-19 C
r1: distance to the first particle = 34.0cm/2 = 17.0cm = 0.17m
r2: distance to the second charge = 0.17m
Then, the electric field is:
[tex]E_T=\frac{k}{r^2}(q_1-q_2)=\frac{(8.98*10^9Nm^2/C^2)}{(0.17m)^2}(4.00*10^{-9}C-1.55*10^{-9}C)\\\\E_T=761.28\frac{N}{C}[/tex]
The electron moves toward the first charge, because it exerts a more strength electric force.
Next, you use the equation (1) to calculate the acceleration:
[tex]a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(761.28N/C)}{9.1*10^{-31}kg}=1.33*10^{14}\frac{m}{s^2}[/tex]
Next, you can calculate the time for which the particle is at 10.0 cm from the charge 1, by using the following formula:
[tex]v^2=v_o^2+2ax[/tex] (3)
x: distance traveled by the electron = 0.17m - 0.10m = 0.07m
vo: initial speed of the electron = 0 m/s
You replace the values of the parameters in the equation (3):
[tex]v=\sqrt{v_o^2+2ax}\\\\v=\sqrt{2(1.33*10^{14}m/s^2)(0.07m)}=4.31*10^6\frac{m}{s}[/tex]
The speed of the electron when it is at 0.10 m from the charge 1 is 4.31*10^6 m/s
