Physics Lab 2 - In the second simulation we added another...

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Conceptual Questions: 1) 2) No, because E=0 does not mean that V=0. E=-d(V)/d(x) so if E=0, V is considered constant but not necessarily zero. Analysis In the first simulation we were able to demonstrate a point charge . Placed a charge of + 4 units in the center of the graph. Drew electric field lines and equipotential lines on the diagram to show the field it produces. We found that the positive force goes directly outward so the electric field lines go out from the charged particle and the equipotential lines make a 90 o angle with the field lines. If you move outwards from the charged particle the electric potential gets closer to zero.
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Unformatted text preview: In the second simulation we added another charge to the diagram but of the opposite charge. Drew out the electric and equipotential lines again like before. The graph showed that the lines go from the positive charge towards the negative value. Then we placed another positive charged particle 5 units away from the midpoint and determined that the particle would be directed towards the negative charge. Then placed a negative particle at the same spot as the positive particle and determined the negative particle would be attracted to the posistive sphere....
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This note was uploaded on 11/17/2010 for the course HORT 306 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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