Lecture 7 - Battery test Battery test-1 A. B. C. D. E. What...

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attery test-- Battery test 1 What is |V 2 -V 1 | for 9V batteries connected like this? Red represents opper wire copper wire. A. 0V. B. 9V. 8V C. 18V D. 27V E. Can’t tell. V 1 V 2
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attery test-- Battery test 2 What is |V 2 -V 1 | for 9V batteries connected like this? Red represents opper wire copper wire. A. 0V. B. 9V. 8V C. 18V D. 27V E. Can’t tell. V 1 V 2
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attery test-- Battery test 3 What is |V 2 -V 1 | for 9V batteries connected like this? Red represents opper wire copper wire. A. 0V. B. 9V. 8V C. 18V D. 27V E. Can’t tell. V 1 V 2
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attery test-- Battery test 4 What is |V 2 -V 1 | for 9V batteries connected like this? Red represents opper wire copper wire. A. 0V. B. 9V. 8V C. 18V D. 27V E. Can’t tell. V 1 V 2
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etting E from V Getting E from V •Since Δ V=- E•dr, it seems reasonable that if we could , invert the integral, we could get E from Δ V. •If the path of integration is along, say the x-axis, this is very simple: Δ V=- E x (r)dx Æ E x (r)=-dV(r)/dx. •The full result just has to allow for the possibility of any ath in space. We need the “3D derivative”, which is known path in space. We need the 3D derivative , which is known as the gradient : k r V j r V i r V r V r E ˆ ) ( ˆ ) ( ˆ ) ( ) ( ) ( = = r r z y x •This looks bad; but looks can be deceiving. This is the route to the calculation of real E-fields.
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ield lines and Equipotentials Field lines and Equipotentials
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I equipotentials The figure below shows equipotentail surfaces in a cross section of the human heart. Pick the point where E is highest. A. B.
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This note was uploaded on 05/11/2010 for the course PHY 214 taught by Professor Timothybolton during the Spring '10 term at Kansas State University.

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Lecture 7 - Battery test Battery test-1 A. B. C. D. E. What...

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