electrostatics

VITEEE 2015 Help › Electricity and magnetism questions › electrostatics

Identical thin rods of length 2a carry equal charges,
Q, uniformly distributed along their lengths. The
rods lie along the x axis with their centers separated by
a distance of (in Fig. ). Show that the magnitude
of the force exerted by the left rod on the right
one is given by

F = (kQ² / 4a²) ln ( b² / b² - 4a² )

#Physics #Electricity and magnetism

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2 Answers

1

Euclid ·2010-11-04 00:32:14

I wish we also had Centre of Charge...:)

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Euclid ·2010-11-04 07:15:20

find field due to the left rod at a distance r from its centre...

so E = âˆ«kÎ»dx2a(r-x)², where Î» = Q/2a

integrating from -a to +a,

E = kQr² - a²

So now coming to question the force on the right rod due to first,

F = âˆ«(kQr² - a²) Î»dr

now integrating (with r as variable) from b-a to b+a,

F = kQ²4a ( log bb+2a - log b - 2ab )

which gives, F = kQ²4a logb²b² - 4a²

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