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SI123

  • 3 years ago

The electric field between the 2 parallel plates of an oscilloscope is 1.2 x 10^5 V/m. If an electron of energy 2 keV enters at right angles to the field, what will be its deflection if the plates are 1.5 cm long?

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  1. Mashy
    • 3 years ago
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    The deflection angle is required?

  2. SI123
    • 3 years ago
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    deflection angle is not mentioned in the book

  3. Mashy
    • 3 years ago
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    then what is asked? what you mean find deflection??

  4. SI123
    • 3 years ago
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    |dw:1361792988295:dw| Y is the deflection that has to be calculated

  5. DLBlast
    • 3 years ago
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    the angle should be 37

  6. DLBlast
    • 3 years ago
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    |dw:1361855358005:dw|for angles 180=x+y+z

  7. DLBlast
    • 3 years ago
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    oh 45 is the other number it could be

  8. Shadowys
    • 3 years ago
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    in this case, since it is constant, uniform acceleration due to the electric field, by Newton's second law, \(\Sigma \vec F= q \vec E = m \vec a\) where q and m of the electron are known constants. so, \(\vec a=\frac{q \vec E}{m}\), which will be different in direction of the electric field, as expected, now, the original speed of the electron can be calculated from: \(\frac{1}{2}mu^2= 2KeV\), so \(u=\sqrt{ \frac{4KeV}{m}}\) by the equations of motion, \(\vec S_x=\vec u_x t + \frac{1}{2} \vec a_x t^2\), but \(\vec u_x=\vec u\) and \(\vec a_x =0\) and \(\vec S_x =1.5cm\) so, \(t=\frac{1.5}{\sqrt{ \frac{4KeV}{m}}}\) now, for \(\vec S_y=\vec u_y t + \frac{1}{2} \vec a_y t^2\), but this time,\(\vec u_y=0\), \(\vec a_y = \vec a = \frac{q \vec E}{m}\) so, \(\vec S_y= \frac{1}{2} \vec a_y t^2\) sub it all in to get the displacement along the y axis. Do you need further help?

  9. DLBlast
    • 3 years ago
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    Cartesian coordinate system + newton law+Gauss's Law+Coulomb's Law=electric field

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