ParticleInFieldPhysletSolution

ParticleInFieldPhysletSolution - € To get the electric...

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Unformatted text preview: € To get the electric field, you first need the acceleration. The acceleration leads to the force, and the force leads to the electric field. Each of these quantities is a vector, so you need to do components. To get the acceleration, you need the velocity at two points, since: v −v ax = fx ix and similarly for the y‐acceleration. t f − ti Any two points in the electric field will work – even those separated by just one time step. I chose two points that were not near the edge of the field, since it wasn’t clear exactly where it began and ended. v ix = 13.5 cm µs t i = 1.75µs v fx = 15.0 µs cm ax = 15.0 − 13.5 cm µs 1.8 − 1.75µs cm = 30 µs2 € t f = 1.8µs Looking at the y­component of velocity, I see it has the same change between those times, € so the same acceleration. It was a bit bothersome to see how many people had trouble dealing with the units. 2 1m 10 6 µs 11 cm ax = 30 µs2 ⋅ ⋅ = 3 × 10 sm 2 100cm 1s Now, ∑ Fx = max qE x = max −31 11 m € max 9.11 × 10 kg ⋅ 3 × 10 s2 N Ex = = = 1.7 C −19 q 1.6 × 10 C The y‐component is equal, and the whole magnitude comes from the Pythagorean theorem (2.4 N/C). € ...
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