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Understanding on potential near a charge |
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| Jun15-12, 02:21 AM | #1 |
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Understanding on potential near a charge
From the fig 1 & 2 in the attachment.
i) what is the sign of potential difference in fig.1 Vp – VQ ii) And in fig.2 VQ – VP For simplicity, lets take value at P,Q and R as in the figure(attach) for a unit +ve charge. Both the answer are positive. I can understand i) is positive but not ii), as that means VQ > VP. I know this is what we get when we use the formulae V = Q/4ε°r. But my understanding near the source charge, the potential to do work should always be greater i.e Vp > VQ . Electric potential at a place is defined as the potential of a unit +ve charge to do work when placed at that point and therefore point near the source would always have greater potential. What is wrong on my understanding? Also, can't we take unit -ve charge as test charge, since it too has the same potential to do work as the unit +ve charge, only direction of work done is opposite. Thanks. |
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| Jun15-12, 06:59 AM | #2 |
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The sign of the potential depends on your reference point. |
| Jun15-12, 08:00 AM | #3 |
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| Jun15-12, 10:25 AM | #4 |
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Understanding on potential near a charge |
| Jun15-12, 11:50 AM | #5 |
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I'm little confused with work done still.
Like when you say work is done ON THE TEST CHARGE - is this work done on the test charge BY AN EXTERNAL AGENT in pushing it against the electric field. And work done BY THE TEST CHARGE - is this work done on the test charge DUE TO SOURCE CHARGE i.e either repellsion or attraction. |
| Jun15-12, 01:19 PM | #6 |
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I think you need to re-examine the relationship between work and energy then revisit the electrostatic case.
http://hyperphysics.phy-astr.gsu.edu...ic/elewor.html The potential is defined as the work (per unit charge) done against the electric field. Thus changes with the field - in the direction the test charge wants to go by itself - would show up as negative. Note: scalar's can be negative numbers. The negative sign does not have to mean direction. |
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