1. The motion of a singly ionized negatively charged ion moving parallel to the
ID: 1586033 • Letter: 1
Question
1. The motion of a singly ionized negatively charged ion moving parallel to the x axis is tracked. At x=5 cm the particle has 1.28 x 10-16 J of kinetic energy at x=8 cm the kinetic energy is
1.0 x 10-16 J at x=11 cm, 8.8 x 10 -17 J and at x=14 cm, 8.4 x 10-17 J.
a) Is there an electric field present? If so, in which direction does the field’s x component point?
Explain your reasoning.
b) Is the field uniform? How can you tell?
c) Field present or not, at each of the four positions the electric potential can be measured. Devise two sets (four measurements each) of electrical potential values that are consistent with the data given.
2. The polar molecule in the diagram below has a positive charge of +19e at one end and a negative charge of -19e at the other. It is immersed in a non- uniform electric field (and aligned with the field direction). The magnitude of the field is 4.6 x 107 N/C at the negative end of the molecule and drops by 3% at the positive end.
a) Determine the net force that acts on the molecule (calculate the magnitude and describe the direction).
b) What would the net force on the molecule be if the field were uniform with a constant magnitude of 4.6 x 107 N/C?
c) Water is a polar molecule. Imagine a stream of water falling through a region of space where an electric field exists. How could this experiment be used to determine if the electric field is uniform or non uniform? Explain your reasoning and describe how you could use your observations to reach a conclusion.
3) The electric field between a charged Van de Graaf generator and its grounding sphere is generally non uniform. This is true in the example diagrammed below where the surface of the generator is at a potential of 5000 V relative to the grounding sphere which is at a potential of 0 V. The shapes of several equipotential surfaces are sketched as well as the velocity vector of an electron that is crossing the 1200 V equipotential surface at a speed of 5.45 x 106 m/s.
a) Determine the change in the electron’s potential energy that will occur if the electron moves from the 2500 V equipotential surface to the surface of the generator.
b) How much kinetic energy will the electron have as it reaches the surface of the generator from the 1200 V surface assuming it doesn’t collide with anything along the way?
c) How much work did the electric field do on the electron during the described motion?
Explanation / Answer
(a)
We can Clearly see, Kinetic Energy is decreasing as move in the +ve x axis.
Yes, there is presence of electric field and it is in the direction in which kinetic energy is decreasing i.e in positive X-direction.
(b)
No, it is not uniform as the kinetic energy is reducing at a non-uniform rate.
Because, K.E = (q*E) * d
(c)
At x = 5 cm K.E = 1.28 * 10-16 J
Potential Energy = q*V
Loss in Kinetic Energy = Gain in Potential Energy
U = q*V = 1.28 * 10-16 J
V = (1.28 * 10-16)/(1.6*10^-19) J/C
V = 800 Volt
At x = 8 cm, K.E = 1.0 * 10-16 J
q*V = 1.0 * 10-16 J
V = (1.0* 10-16)/(1.6*10^-19) J/C
V = 625 Volt
At x = 11 cm, K.E = 8.8 * 10^-17 J
q*V = 8.8 * 10^-17 J
V = (8.8* 10^-17)/(1.6*10^-19) J/C
V = 550 Volt
At x = 14 cm, K.E = 8.4 * 10^-17 J
q*V = 8.4 * 10^-17 J
V = (8.4 * 10^-17)/(1.6*10^-19) J/C
V = 525 Volt
Please post seperate questions in seperate post.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.