Introductory Physics Volume Two

178 Hints A
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Hints
1.1 Do the directions “by hand”. To make it easier to keep track of
everything write all forces in terms of the quantity F 0 =
1.2 Use the vector form of Coulomb’s law.
1.3 The answer is E(x, ⃗ 0) = q
4πɛ 0
−2aĵ
(x 2 +a 2 ) 3/2 .
q2
4πɛ 0a
. 2
1.4 Use polar coordinates, ⃗r s = R cos θî + R sin θĵ and dq = Q 2π dθ.
1.5 Use Gauss’s Law.
1.6 Do not attempt to integrate the flux over the surfaces. First,
look carefully at the orientation of the electric field through each of the
three faces that touch the charge. Second, because of the symmetry of
the configuration the remaining three faces must have the same flux as
each other. Third, notice that if you placed eight such cubes around
the charge (forming a larger cube with the charge at the center), that
all eight smaller cubes would have the same flux.
1.7 Use a gaussian surface that is a sphere of radius r, centered on
the charge.
1.8 Use Gauss’s law. The gaussian surfaces are spheres. If r < R
then the amount of charge inside the gaussian surface depends on r,
show that the amount of charge inside is q in = Q r3
R 3
1.10 Assume that the field is directed straight out from the line. Let
the Gaussian surface be a cylinder (like a tin can) with the line charge
as the axis of the cylinder. Note that the flux through the ends of the
can is zero because of the orientation relative to the field.
1.11 The gaussian surface is a cylinder of radius r and length L.
1.12 Recall that there can be no field inside the body of a conductor
that is in static equilibrium.
1.13 The geometry gives you the angle of the forces, but because
the charges are not of the same size you must still deal with both
components. Depending on your disposition, you might prefer using
the vector form of Coulombs law.