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Friday, March 22, 2019

An electron moves between two points X and Y in a uniform electric field, as shown.


Question 11
An electron moves between two points X and Y in a uniform electric field, as shown.


The distance between X and Y is 4.0 cm and the line XY is at an angle of 60° to the direction of the field.

The field exerts the only force on the electron.

The field strength is 100 N C-1.

What is the change in the kinetic energy of the electron as it moves from X to Y?
A – 4 eV                      B –2 eV                       C +2 eV                       D +4 eV





Reference: Past Exam Paper – November 2017 Paper 12 Q32





Solution:
Answer: B.

Electric force F = Eq

Work done (change in energy) = Fd = Eqd
where d is the distance moved in the direction of the force


XY = 4.0 cm at 60° to the field
Distance moved in direction of force = XY cos 60° = (4×10-2 × cos 60°) m


Change in energy = Fd = 100 × e × 4×10-2 × cos 60°
Change in energy = 2 eV

The energy is in electron-volt (eV) as the value of the charge of the electron has not been substituted.
 

The direction of the force on an electron is opposite to the direction of the field, so the electric field in this case decelerates the electron. [B is correct]

2 comments:

  1. Why did you use cos and not sin(60) or tan(60) ??

    ReplyDelete
    Replies
    1. because we need the component of XY ALONG the field

      follow the instructions above and draw on the diagram. you will observe that to obtain the horizontal component, we need to use cos 60

      Delete

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