dy y L/2 -L/2 r D X

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A charge Q=2.85×10−4 C is distributed uniformly along a rod of length L, extending from =−27.7/2 cm to =+27.7/2 cm, as shown in the diagram below. A charge q=6.3×10−6 C, and the same sign as Q, is placed at (D,0), where D=20.5 cm.

Use integration to compute the total force on q in the x direction.

When I do this problem I havegotten .7395 N, 15.161 N, .046 N, .271 N all of which are wrong.Can you please help

dy=
y
L/2
-L/2
r
q
D
X
Transcribed Image Text:dy= y L/2 -L/2 r q D X
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