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Unformatted text preview: 25. Consider an inﬁnitesimal section of the rod of length dx, a distance x from the left end, as shown in the
diagram below. It contains charge dq = λ dx and is a distance r from P . The magnitude of the ﬁeld it
produces at P is given by
1 λ dx
.
4πε0 r2
1 λ dx
sin θ
The x component is dEx = −
4πε0 r2
1 λ dx
cos θ .
and the y component is dEy = −
4πε0 r2
dE = y dq x R .
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.. x r θ .
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.. P
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.... d E
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... We use θ as the variable of integration and substitute r = R/ cos θ, x = R tan θ and dx = (R/ cos2 θ) dθ.
The limits of integration are 0 and π/2 rad. Thus,
Ex = − π /2 λ
4πε0 R sin θ dθ =
0 and
Ey = − λ
4πε0 R λ
cos θ
4πε0 R π /2
0 =− λ
4πε0 R λ
sin θ
4πε0 R π /2
0 =− λ
.
4πε0 R π /2 cos θ dθ = −
0 We notice that Ex = Ey no matter what the value of R. Thus, E makes an angle of 45◦ with the rod
for all values of R. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge

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