Force = 202,200 x 1.5 ms.q
Force = 303,000 newtons
The final speed of the electron is 4.64 * 10^5 m/s.
<h3>What is the speed of the electron?</h3>
Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
Learn more about the speed of the electron:brainly.com/question/13130380
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Average acceleration is
Change in Velocity/change in time
So you could then do Vf-Vi/Tf-Ti
Which would look like 13m/s-6m/s / 1s-0s
Which then is 7m/s/1s which means the acceleration is 7m/s^2
Answer:
Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?
Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?
