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Prove this result Electrostatics

  1. Jun 6, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A ball of mass m1 and charge q1 is projected at an angle ? in a uniform electric field for some time T. The direction of the velocity of the ball changes by 60° and the magnitude of velocity is reduced to half. After removing this ball second ball of mass m2 and charge q2 is projected at the same angle ? and for time T. The initial velocity of both the balls is same. The direction of velocity of second ball changes by 90°. Prove that (q1/m1)/(q2/m2)=3/4

    2. Relevant equations

    3. The attempt at a solution
    I assume the electric field to be in the +ve X direction as Ex and in the +ve Y direction as Ey.

    For q1
    [itex]v_x = u_0 cos \theta + k_1 E_x T=u_0/4 \\
    v_y=u_0 sin \theta + k_1 E_y T= \sqrt{3}u_0/4 [/itex]

    For q2
    [itex]v_x = u_0 cos \theta + k_2 E_x T=0 \\
    T=\dfrac{-u_o cos \theta}{E_x k_2} [/itex]

    But there are too many variables and its seems impossible to me to get rid of them.
     
  2. jcsd
  3. Jun 6, 2013 #2

    mfb

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    What is meant with the angle "?"?

    I guess that gravity is not considered, and that you chose x to be the direction of the initial motion (write that down please).
    What is θ?

    Did you draw a sketch? Initial motion, additional motion due to the electric field, resulting motion?
     
  4. Jun 6, 2013 #3
    I would not use coordinates in this problem. Staying in the vector formalism leads to the solution quite naturally.
     
  5. Jun 6, 2013 #4

    utkarshakash

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    That was a typing mistake. The "?" means θ.
     
  6. Jun 6, 2013 #5

    utkarshakash

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    Can you give me some hints?
     
  7. Jun 7, 2013 #6
    A body with mass ##m## and charge ##q## and initial velocity ##\vec{u}## spending time ##T## in uniform electric field ##\vec{E}##, leaves it with velocity ##\vec{v}##. Newton's second law: ## m\vec{v} = m\vec{u} + qT\vec{E}##, or ## \vec{v} = \vec{u} + k \vec{p}##, where ## k = \frac q m ## and ## \vec{p} = T\vec{E}##. Now you can use the dot product to find out the magnitude of ##\vec{u}## and its angle with ##\vec{v}##.
     
  8. Jun 7, 2013 #7

    utkarshakash

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    Dot product of which two vectors??
     
  9. Jun 7, 2013 #8
    Depends on what you want to find out. Recall the equation relating the dot product with the magnitudes of and the angles between vectors.
     
  10. Jun 7, 2013 #9

    utkarshakash

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    In the question what is meant by the term 'changes by 60'? Does it mean that the angle which the final velocity vector makes with the horizontal is 60° or is it just the angle between initial and final vectors?
     
  11. Jun 7, 2013 #10
    Why would that be "horizontal", and not "vertical" or perhaps some other fixed direction?

    For me, "changes by" implies that the comparison is made between the initial and the final velocities.
     
  12. Jun 7, 2013 #11

    utkarshakash

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    Multiplying u to both sides of the equation you just gave me I get
    [itex]k_1 = \dfrac{\vec{v}.\vec{u}-1}{\vec{p}.\vec{u}} \\
    k_2 = \dfrac{\vec{v'}.\vec{u}-1}{\vec{p}.\vec{u}} [/itex]

    Dividing both equations
    [itex]\frac{k_1}{k_2} = \dfrac{\vec{v}.\vec{u}-1}{\vec{v'}.\vec{u}-1}[/itex]

    Now the dot product of v' and u is zero and that of v and u is u^2/4. But plugging these data won't give me the answer
     
  13. Jun 7, 2013 #12
    You seem to assume that ## \vec {u} \cdot \vec {u} = 1 ##. Why?
     
  14. Jun 7, 2013 #13

    utkarshakash

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    Oh! I later realised that it would be u^2. It was a silly mistake from my side. Anyway, thanks for helping.
     
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