Clayton State
University
Department =
of Natural
Sciences
June 3, 2008<=
/p>
Physics 111=
2 –
Quiz 2
=
=
Name
_____SOLUTION________________________________
=
1. Particles of charge -20.0 mC, +35.0 mC and - 56.0 =
mC are placed
in a line as shown below. The center one is 0.500 m from each of the others.
Calculate the net electric field due to the charges 0.500 m to the right fr=
om
third charge.
![](3D"Summer08.Phys1112.Quiz2.SolutionWeb_files/image002.jpg")
=
q1
=3D -20.0 x 10-6 C
=
q2 =3D +35.0 x 10-6 C
=
=
q3
=3D -56.0 x 10-6 C
=
E1 =3D k | q1|/r12
=3D (8.99 x 109 N-m2/C2) (20.0 x 10-6=
C)/(1.50 m)2 =3D 0.799 x 105 N/C,
to the left
E2 =3D k |q2 |/r22
=3D (8.99 x 109 N-m2/C2) (35.0 x 10-6=
C)/(1.00 m)2 =3D 3.14 x 105 N/C,
to the right
E3 =3D k | q3|/r32
=3D (8.99 x 109 N-m2/C2) (56.0 x 10-6=
C)/(0.500 m)2 =3D 20.1 x 105 N/C,
to the left
Etot
=3D E1 + E2 + E3 =3D -0.799 x 105 N/C +
3.14 x 105 N/C - 20.1 x 105 N/C =3D -1.78 x 106 N/C (Net elect=
ric
field points to the left.)
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