Why Doesn't Calculating Chord Slope Work for Electron Deflection Angles?

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Calculating the deflection angle for an electron in a parallel plate capacitor requires using the final vertical and horizontal velocities rather than just the distances traveled. The initial method of finding the angle based on horizontal and vertical distances leads to incorrect results because it approximates the slope of the chord rather than the tangent at the exit point. The tangent slope at exit is typically about twice the slope of the chord. Understanding this distinction is crucial for accurate calculations in physics problems involving electron deflection. Clarifying the difference between chord and tangent slopes can help prevent similar errors in future calculations.
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When asked to find the deflection angle for an electron passing through a parallel plate capacitor, for some reason my initial approach was to find the horizontal distance traveled and the vertical distance and then find the angle created by them. I got the wrong answer doing this and looked at the solution which explained that you have to use the final vertical/horizontal velocities and the angle created by them. This makes sense to me, but I still don't understand why my method wouldn't work? Any clarification would be appreciated. Thanks!
 
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Hi maccha! :smile:

It's because you're finding the slope of the chord from entry to exit (which will be roughly parallel to the tangent at half-way), but the question asks for the slope of the tangent on exit, which is about twice as much. :wink:
 

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