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Homework Statement
Two thin cylindrical shells, respectively with radius ##R_1 = 12 cm## and ##R_2 = 20 cm## and height ##D = 100 m##, are concentric and characterized by a charge density ##\lambda_1 = -0.38 \frac{\mu C}{m}##, for the internal cylinder, and ##\lambda_2 = 0.32 \frac{\mu C}{m}##, for the external one. An electron (##m_e = 9.1 \cdot 10^{-31} kg## and a charge ##e = 1.6 \cdot 10^{-19} C##) is released stationary from the internal cylinder's surface; what is the velocity when it reaches the other cylinder?
Homework Equations
Gauss Teorem
##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{Q}{\epsilon_0}##
and
##a = \frac{q}{m}E##
The Attempt at a Solution
So, I started by calculating the electric field on both the cylinders.
##\Phi_S (\vec E_1) = \int_S \vec E_1 \cdot d\vec S = E_1 2 \pi r D = \frac{\lambda_1 D}{\epsilon_0}##
so
##E_1 = \frac{\lambda_1}{2 \pi r \epsilon_0}##
Same thing for the other one.
##E_2 = \frac{\lambda_2}{2 \pi r \epsilon_0}##
At this point I should calculate the velocity starting from the acceleration ##a = \frac{e}{m_e}E##, but I'm blocked here. The ##E## in the acceleration is the "total" electric field or it is ##E_2## since the electron is now moving inside the external cylinder?
I know there is this other equation:
##v = \sqrt{\frac{2q(V_1 - V_2)}{m}}##
But in this case I should calculate the two electric potential, which would be:
##\int_A^B \vec E_0 \cdot d\vec l = V(A) - V(B)##
with ##A## being the internal cylinder and ##B## the other one.
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