An electron moving parallel to the x axis has an initial speed of 5 × 106 m/s at

Question

An electron moving parallel to the x axis has
an initial speed of 5 × 106 m/s at the origin.
Its speed is reduced to 4×105 m/s at the point
xP , 2 cm away from the origin. The mass of
the electron is 9.10939 × 10-31 kg and the
charge of the electron is -1.60218 × 10-19 C.
Calculate the absolute value of the poten-
tial difference between this point and the ori-
gin.
Answer in units of V.
013 (part 2 of 2) 10.0 points
Consider the setup from part one. Let U0 and
V0 denote the potential energy and the electric
potential at the origin. Let UP and VP denote
the potential energy and the electric potential
at the final point xP of the electron.
Which of the following statements is true?
1. VP < V0 and UP > U0
2. VP = V0 and UP < U0
3. VP > V0 and UP < U0
4. None of these.
5. VP = V0 and UP = U0
6. VP < V0 and UP = U0
7. VP = V0 and UP > U0
8. VP > V0 and UP > U
9. VP < V0 and UP < U0
10. VP > V0 and UP = U0

Explanation / Answer

  Work done = Change in kinetic energy   q V = (1/2) m ( vf^2 - vi^2 ) (- 1.60218 x 10^-19 ) * V = 0.5 * 9.10939 * [ ( 4 x 10^5)^2 - (5 x 10^6)^2 ]   So, V = 70.62 V   Electron moves from high to low potential during deceleration. Here, there is an increase in potential energy with a corresponding decrease in kinetic energy.   True statement: Option - 1 : VP < V0 and UP > U0

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