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2
particle
0
2
arc
0
/4
2 sin( / 2)/ 4
2sin( / 2)
E
QR
EQ
R
πε
θ
θπ
ε
==
.
With
= π
, we have
particle
arc
1.57.
2
E
E
π
=≈
31. First, we need a formula for the field due to the arc.
We use the notation
λ
for the
charge density,
λ
=
Q
/L
.
Sample Problem 223 illustrates the simplest approach to
circular arc field problems. Following the steps leading to Eq. 2221, we see that the
general result (for arcs that subtend angle
) is
[]
arc
00
2s
i
n
(/
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This note was uploaded on 06/07/2010 for the course PHYS 344 taught by Professor Bb during the Spring '10 term at The Petroleum Institute.
 Spring '10
 BB
 Physics, Charge

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