Physics problems?

Question

a) The anode (+ terminal) of an X-ray tube is at a potential of +119000V with respect to the cathode (- terminal). How much work (J) is done by the electric force when an electron is accelerated from the cathode to the anode? Also if the electron is initially at rest, what kinetic energy does the electron have when it arrives at the anode?

b) The potential at location A is 440V. A positively charged particle is released from rest & arrives at location B with a speed vB. The potential at location C is 784V, & when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2vB. Find the potential at B.

c) Two hollow metal spheres are concentric with each other. The inner sphere has a radius of 0.1740m & a potential of 86.0V. The radius of the outer sphere is 0.176m & its potential is 84.0V. If the region between the spheres is filled with Teflon, find the electric energy contained in this space.

Answer 1

a) energy = qV = 1/2mv^2

so v = sqrt (2qV/m). They give you V. Look up q and m of an electron.

b) vb = sqrt (2q (VA-VB))
vc = sqrt (2q (VC-VB)) = 2vc = 2 sqrt (2q (VA-VB))

(VA-VB)*4 = (VC-VB)

VB = (4VA - VC) / 3

c) for a thin gap, E = (V2-V1)/(r2-r1)

energy = volume * 1/2 k epsilon0 E^2

You'll have to look up the dialectric constant of teflon (k) and epsilon0.

Answer 2

a) W = Vq = 119000 (1.602×10^-19) = 1.91×10^-14 J. This is also the amount of kinetic energy the electron will have when it reaches the anode.

b) W = Vq = (1/2) m v^2.
When released from C, the particle has twice the speed and hence four times the kinetic energy compared to when it was released from A. It follows that the potential difference between C and B must be four times to potential difference between A and B. So we get
784 - V_B = 4 (440 - V_B)
=> 3V_B = 976
=> V_B = 325V.

c) According to the reference below, the dielectric constant of Teflon is 2.1, so the permittivity is ε = 2.1ε_0 = 1.86×10^-11 F/m
and the electric field strength is 2.0 / 0.0020 = 1.0×10^3 V/m
so the energy density is (1/2)(1.86×10^-11)(1.0×10^3)^2 = 9.30×10^-6 J/m^3.
The volume is (4π/3) (0.176^3 - 0.1740^3) = 1.84×10^-4 m^3, so the total energy stored is
9.30×10^-6 (1.84×10^-4) = 1.71×10^-9 J.