A21-Lect-6+%5BCompatibility+Mode%5D

# A21-Lect-6+%5BCompatibility+Mode%5D - The electric field of...

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1/25/2009 1 ± The electric field of a charged disk ¾ The figure below shows a disk of radius R having total charge Q which is uniformly distributed on the surface. ¾ The surface charge density is 2 R Q A Q π η = = ² Our objective is to calculate the on-axis electric field. ¾ Following the problem solving strategy, we need to divide the continuous charge of the disk (total charge Q) into N very narrow rings of charge Q . Because we’ve already learn how to find the electric field of a ring. ¾ As shown in the fig the radius of ring i is r i , has a width r, area A i and charge Q i . ¾ The electric field of ring i, at a point along the z-axis is given by, ¾ The on-axis electric field of the whole disk is the sum of the electric field of all of the rings: 2 / 3 2 2 0 ) ( 4 1 ) ( i i z i r z Q z E + = = = N i N z i z disk Q z E E 2 3 2 2 ) ( ) ( + = 2 / 3 2 2 0 ) ( 4 1 ) ( R z zQ E z ring ¾ Since we deal with a surface rather than a line. Thus the charge Q on each ring is ¾ The area A i can be found by “ unrolling the ring to form a narrow rectangle of length 2 π r ( the perimeter ) and width r”. = = + i i i r z 1 / 0 1 ) ( 4 i i A Q A Q = = i ¾ Thus, the area of ring i is, ¾ Thus, r r Q Thus r r A i i i = = 2 , 2 = + = N i i i z disk r z r r z E 1 2 / 3 2 2 0 ) ( 2 ) (

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1/25/2009 2 ¾ We want to let then the width . Now we can replace the sum with an integral ranging from . Now we can write the electric field as, ¾ If we change the variable as N dr r R r to r = = 0 + = R i z disk r z rdr z E 0 2 / 3 2 2 0 ) ( 2 ) ( η If we change the variable as, ¾ Using the new variable, the lower integration limit is ¾ And the upper limit is Note: in a definite integral, if you change the variables, then you must du rdr rdr du r z u 2 1 2 2 2 = ⇒∴ = + = 2 0 z u r = = 2 2 R z u R r + = = change the limits. ¾ The electric field with new variables becomes, + = = = + + 2 2 0 2 / 1 0 z z 2 / 3 0 1 1 2 2 4 2 1 2 ) ( | 2 2 2 2 2 2 R z z z u z u du z E R z z R z disk ¾ If we multiply and divide the R.H.S of the equation by z we get , ± Now, let’s check the limiting case (an approximation).
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A21-Lect-6+%5BCompatibility+Mode%5D - The electric field of...

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