If you have a discrete charge distribution 1 2 3

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If you have a discrete charge distribution (1, 2, 3... charges) then you can add up their individual scalar potentials. The electric potential for this differential charge contribution will be given by: dV = k e dq r To find the total electric potential from the entirety of the distribution, then sum over all the differential contributions. V = dV ! = k e dq r !
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+ + + + + + + + + + Continuous Distributions For example, let’s calculate the electric potential at a point P which is a distance z along the central axis due to a charged ring of radius R and uniform linear charge density, λ (total charge q). Solution : dV ds dq z P A small segment of wire ds has a charge of dq= λ ds. R At point P, the differential electric field contribution will be: r = z 2 + R 2 dV = k e dq r dV = k e ! ds z 2 + R 2
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Continuous Distributions Integrate to get: ds dq z z P We can check to see if this matches our previous E field calculation: + + + + + + + + + + R r = z 2 + R 2 V = dV ! = k e ! ds z 2 + R 2 ! V = k e 1 z 2 + R 2 ! ds ! V = k e q z 2 + R 2 dV E z = ! " V " z E z = ! " " z k e q z 2 + R 2 # $ % % & ' ( ( E z = k e qz z 2 + R 2 ( ) 3/2 Same as before
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For Next Time (FNT) Finish the homework for Chapter 24 Start reading Chapter 25
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