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An electron moving parallel to the x axis has an initial speed of 5.38 x 106 m/s at the origin. Its speed is reduced to 1.68 x 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. -0.07425 x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Volts (b) Which point is at the higher potential? O the point x = 2.00 cm O both have the same potential o the origin

An electron moving parallel to the x axis has an initial speed of 5.38 x 106 m/s at the origin. Its speed is reduced to 1.68 x 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. -0.07425 x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Volts (b) Which point is at the higher potential? O the point x = 2.00 cm O both have the same potential o the origin

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An electron moving parallel to the x axis has an initial speed of 5.38 x 106 m/s at the origin. Its speed is reduced to 1.68 x 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. -0.07425 x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Volts (b) Which point is at the higher potential? the point x = 2.00 cm both have the same potential o the origin Need Help? Read It Master It
Transcribed Image Text:An electron moving parallel to the x axis has an initial speed of 5.38 x 106 m/s at the origin. Its speed is reduced to 1.68 x 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. -0.07425 x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Volts (b) Which point is at the higher potential? the point x = 2.00 cm both have the same potential o the origin Need Help? Read It Master It
Expert Solution
Step 1

Part (a)

Given:

  • The initial speed of electron is u=5.38×106 m/s.
  • The final speed of electron is v=1.68×105 m/s.
  • The distance covered by electron is x=2.00 cm.

The formula to calculate the change in kinetic energy of the electron is,

∆K=12mv2-u2

Here, ∆K is the change in kinetic energy, m is the mass, v is the final speed and u is the initial speed.

The formula to calculate the change in potential energy of the electron is,

∆U=q∆V

Here, ∆U is the change in potential energy, q is the charge on electron and ∆V is the potential difference.

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