Physics notes electric field PDF

Title Physics notes electric field
Course Modern Physics
Institution Butler University
Pages 2
File Size 36.5 KB
File Type PDF
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Summary

Physics assignments, study guides, and notes weeks 6-9...


Description

Physics notes electric field Electric field To help visualize how a charge, or a collection of charges, influences the region around it, the concept of an electric field is used. The electric field E is analogous to g, which we called the acceleration due to gravity but which is really the gravitational field. Everything we learned about gravity, and how masses respond to gravitational forces, can help us understand how electric charges respond to electric forces. The electric field a distance r away from a point charge Q is given by: Electric field from a point charge : E = k Q / r2 The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. Like the electric force, the electric field E is a vector. If the electric field at a particular point is known, the force a charge q experiences when it is placed at that point is given by : F = qE If q is positive, the force is in the same direction as the field; if q is negative, the force is in the opposite direction as the field. Learning from gravity Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 m/s2 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 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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