ch15-3-2
TRANSCRIPT
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What is the
direction of the
electric field at
the observation
location Adue to
piece 2?2
A
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What is the
direction of the
electric field at
the observation
location Adue to
piece 7?7
A
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What is the
direction of the
electric field at
the observation
location Adue
to the two
pieces 2 and 7? 72
A
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What is the
direction of the
electric field at
the observation
location Adue
to all eight
pieces? 72
A
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What is the
direction of the
electric field at
the observation
location Aif the
rod has a
uniform charge
of -Q?
72
A
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A rod of length L has
a uniform charge of
+Q. The rod is
divided into equal
pieces of length x
as shown. What is
the charge Q on one
slice?
1)QL/x2)Qx/L3)Q(L-x)/L4)2QL/x
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What are the
limits of
integration?
1)0 to 2)to +3)L to + L4)L/2 to +L/25)0 to L
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Which will give
us a more
exact result:
1)Dividing the rod into a few slices and adding upthe electric field of each slice by hand.
2)Dividing the rod into many slices and adding upthe electric field of each slice using a computer.
3)Dividing the rod into infinitesimally small slicesand adding the electric field of each slice using
integration.
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Q15.1aHere is a rod lying on the y axis. It is
uniformly charged positive.
What is the direction of the net
electric field at location A?
A
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Q15.4aUniformly charged ring of charge Q andradius R
What is Q, expressed in terms ofgiven constants and an integrationvariable?
1) Q /
2) Q/ (2R)3) Q* / (2R)
4) Q* / 2
5) Q*R*/(2R)
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Q15.4bUniformly charged ring of charge Q andradius R
What are the integration limits?
1) 0 to
2) to +3) 0 to
4) 0 to2
5) 2to +2
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Q15.4cBy thinking about the physicalsituation, predict the magnitude ofthe electric field at the center of auniformly charged ring of radius Rcarrying a charge +Q. Then use theformula we just derived to confirm
this result.
1) E = 1/(40) Q/z2
2) E = 1/(40) Qz/R3
3) E = 0
4) E =
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Q15.7aThe electric field at any location inside a hollow, uniformly charged sphere,due to the charge on the surface of the sphere, is zero. True or false?
1) True, because the electric field can't penetrate the material of the sphere2) True, because the contributions of all the charges on the surface add up
to zero at every location inside the sphere.
3) True, but only at the center of the sphere, where all the contributions cancel.4) False, because the electric field due to the surface charges penetrates
the sphere.