Answer:
[tex]v=8.89\times 10^{15}\ m/s[/tex]
Explanation:
An electron is released at rest from point A. It moves in response to the electric field of a fixed charge distribution along a path that takes it through point B.
Potential at point A, [tex]V_A=-11\ V[/tex]
Potential at point B, [tex]V_B=25\ V[/tex]
Let v is the speed of the electron as it passes point B. It can be calculated using the conservation of energy theorem as :
[tex]\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=V_B-V_A[/tex]
Here, u = 0
[tex]\dfrac{1}{2}mv^2=25-(-11)[/tex]
[tex]mv^2=72[/tex]
[tex]v=\sqrt{\dfrac{72}{m}}[/tex]
m is the mass of electron
[tex]v=\sqrt{\dfrac{72}{9.1\times 10^{-31}}}[/tex]
[tex]v=8.89\times 10^{15}\ m/s[/tex]
So, the electron's speed as it passes point B is [tex]8.89\times 10^{15}\ m/s[/tex]. Hence, this is the required solution.
