If An Electron (m=9.1×10−31kg) Escaped From The Surface Of The Inner Cylinder With Negligible Speed, What Would Be Its Speed When It Reached The Outer Cylinder?
A thin cylindrical shell of radius R1=6.5cm is surrounded by a second cylindrical shell of radius R2=9.3cm as in the figure (Figure 1) . Both cylinders are 15 m long and the inner one carries a total charge Q1=−0.77μC and the outer one Q2=+1.54μC.
If an electron (m=9.1×10−31kg) escaped from the surface of the inner cylinder with negligible speed, what would be its speed when it reached the outer cylinder?
First, let's determine the electic field inside between the inner and outer shells. This can be done using Gauss Law:
`int vecE*dvecS = Q/epsilon_0`
Here, Q is the charged enclosed by Gaussian surface. If we choose the Gaussian surface to be a cylindrical shell between the two cylinders (that is, its radius r is
`R_1 < r < R_2` ), then the enclosed charge will be the charge on the inner shell, `Q_1` . Since the shells' radii are much smaller than the their length, we can approximate the electric field as having only radial, and no vertical, component. (Please see the picture on the reference link.)
Using the symmetry considerations, the integral can be rewritten as
`E*2*pi*r*L = Q/epsilon_0` .
From here, the electric field everywhere between the two shells will be
`E = Q/(2*pi*epsilon_0*L*r)` . Since the charge of the inner shell is negative, the electric field vector is directed radially towards the center.
To find the speed of electron when it reaches the outer cylinder, we can use the law of the conservation of energy:
`DeltaK + DeltaU = 0`
The change of kinetic energy of electron is
`DeltaK = (mv_f^2)/2 - (mv_i^2)/2` . Since the initial velocity is negligible,
`DeltaK = (mv_f^2)/2` , where the final speed is the speed where the electron reaches the outer cylinder, which is what we are looking for.
The change in potential energy is
`DeltaU = e*DeltaV` , where `DeltaV` is the change in the electric potential of the electric field we have found. The change of electric potential between inner and outer shells can be found by integrating the opposite of the electric field:
`DeltaV =- int_(R_1) ^(R_2) Edr = -int_(R_1) ^ (R_2) Q/(2*pi*epsilon_0*L*r) = -Q/(2*pi*epsilon_0*L) ln(R_2/R_1)`
Plugging in the given values, we can find that `DeltaV = 331 ` Volts and
`DeltaU = eDeltaV = -1.602*10^(-19) * 331 = -5.3*10^(-17) J`
From the equation of the conservation of energy above,
`DeltaK = (mv_f^2)/2 = -DeltaU = 5.3*10^(-17)` Joules,
and the final velocity of the electron will be
`v_f = sqrt((2*5.3*10^(-17))/(9.1*10^(-31))) = 1.1 *10^7 m/s`
The speed of electron when it reaches the outer shell will be 1.1 *10^7 m/s.
To answer the question about the proton, use the second Newton's Law:
`vecF = mveca`
Since the proton is moving in a circle with a constant speed, its acceleration is directed radially towards center, and is equal
`a = (v^2)/r` . The centripetal force is provided by electric field (since it directed towards center, and proton has a positive charge, it will also be directed towards center) and equal
`F = qE` , where q is the charge of a proton (same as that of electron, but positive), and E is the electric field found previously. Plugging these expressions into the second Newton's Law, we get
`(m_pv^2)/r = q*Q/(2*pi*epsilon_0*L*r)` (Here, consider Q to be positive, because we already took the direction into consideration.)
Notice that radius cancels, and the speed of the proton is
`v = sqrt((qQ)/(2*pi*epsilon_0*L*m_p))`
Plugging in all numerical values, we get v = 9.4*10^5 m/s.
The speed of proton is 9.4*10^5 m/s.