The motion of a singly ionized negatively charged ion moving parallel to the x a

Question

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 fields 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.

Explanation / Answer

(a) With respect to the x-axis there is a moving negative particle with charge -e(-1.6 x 10^-19 C).
Since this particle is moving in such a direction that its KE is getting decreased as a result of electric field present in its path which must be stopping it. Therefore we can say easily that the Electric field direction must be opposite to the movement of the charged particle.i.e it is in -ve x direction.
(b)This field must be uniform , only then there can be a consistent decreased in the Kinetic Energy of the particle. We see that K. E is consistently decreasing
(c)
Difference in KE1(at x= 5cm from rest point)= KE1 = 1.28 x 10^-16 J = Difference in potential energy U1
U1= [e q / 4 (pie) (epsilon) ][(1/r -1/r1)]= 1.28 x 10^-16 J
Similarly,
U2 (at x = 8cm)= [e q / 4 (pie) (epsilon) ][(1/r -1/r2)]= 1.0 x 10^-16 J
On solving both the equations we get,
q = 1.9 x 10 ^-8 C.
Also Potential can be found by
V1(at x= 5cm ) =K.E 1 (at x=5cm) / e ={[ 1/ 4 x (Pie) x (epsilon) ] x q}/x = 800 V
V2(at x= 8cm ) =K.E 2 (at x=8cm) / e = {[ 1/ 4 x (Pie) x (epsilon) ] x q}/x = 625 V
V3(at x= 11cm ) =K.E 3 (at x=11cm) / e = {[ 1/ 4 x (Pie) x (epsilon) ] x q}/x = 550 V
V4(at x= 14 cm ) =K.E 4 (at x=14cm) / e = {[ 1/ 4 x (Pie) x (epsilon) ] x q}/x = 525 V
Answer

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