A thin cylindrical shell of radius R1 = 6.0 cm is surrounded by a second cylindr

Question

A thin cylindrical shell of radius R1 = 6.0 cm is surrounded by a second cylindrilical shell of radius R2 = 8.9 cm as in the figure. Both cylinders are 15 m long and the inner one carries a total charge Q1= -0.81 micro Coulombs and the outer one Q2= + 1.40 microCoulombs

If an electron (9.1*10^-31 kg) escaped from the surface of the inner cylinder with negligible speed, what would be its speed when it reached the outer cylinder?

If a proton (m= 1.67 *10 ^ -27 kg) revolves in a circular orbit of radius r= 6.9 cm about the axis (i.e., between the cylinders), what must be its speed?

Explanation / Answer

a)

By apply Gauss' Law:

E = ?/(2??0R1) >>>>> E = Q1/(2??0rL)

?V = ?E.dr (integration from R1 to R2)

>>>>> ?V = ?[Q1/(2??0rL)].dr

>>>>> ?V = (Q1/(2??0L)) * ln(R2/R1)

>>>>> ?V = (0.81e-6/(2*3.14*8.85e-12*15)) * ln(0.089/0.06)

>>>>> ?V = 382.9 V

Conservation of energy:

(1/2) m v2 = q ?V

>>>>> v = ?(2q(?V)/m)

>>>>> v = ?(2*1.6e-19*382.9/9.1e-31)

>>>>> v = 1.16 * 10^7 m/s

----------------------------------------------------------------------

b)

F = mv2/r = qE >>>> v = ?(rqE/m)

E = Q1/(2??0rL)

E = 0.81e-6/(2*3.14*8.85e-12*6.9e-2*15)

E = 14074.13 V/m

>>> v = ?(rqE/m)

>>> v = ?(6.9e-2*1.6e-19*14074.13/1.67e-27)

>>> v = 3.05 * 10^5 m/s

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