Learning from gravity

Learning from gravity - from the constant force experienced...

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Learning from gravity Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 m/s 2 and points straight down. If you threw a mass through the air, you know it would follow a parabolic path because of gravity. You could determine when and where the object would land by doing a projectile motion analysis, separating everything into x and y components. The horizontal acceleration is zero, and the vertical acceleration is g. We know this because a free-body diagram shows only mg, acting vertically, and applying Newton's second law tells us that mg = ma, so a = g. You can do the same thing with charges in a uniform electric field. If you throw a charge into a uniform electric field (same magnitude and direction everywhere), it would also follow a parabolic path. We're going to neglect gravity; the parabola comes
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Unformatted text preview: from the constant force experienced by the charge in the electric field. Again, you could determine when and where the charge would land by doing a projectile motion analysis. The acceleration is again zero in one direction and constant in the other. The value of the acceleration can be found by drawing a free-body diagram (one force, F = qE) and applying Newton's second law. This says: qE = ma, so the acceleration is a = qE / m. Is it valid to neglect gravity? What matters is the size of qE / m relative to g. As long as qE / m is much larger than g, gravity can be ignored. Gravity is very easy to account for, of course : simply add mg to the free-body diagram and go from there. The one big difference between gravity and electricity is that m, the mass, is always positive, while q, the charge, can be positive, zero, or negative....
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This note was uploaded on 11/22/2011 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.

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