Answers - Electric Fields

1.

The magnitude of the force of +Q or -Q on qo at r = 3 m is FqQ = kQqo/r2 = (9.0 x 109 N-m2/C2)(2 x 10-6 C)(10-12 C)/(9 m2) = 2 x 10-9 N.

The force of +Q on +q0 at

P1 is to the right since +Q repels +qo and urges it to the right

P2 is to the left since +Q repels +qo and urges it to the left.

The force of -Q on + qo at

P1 is to the left since -Q attracts +qo and urges it to the leftAnswers - Electric Fields

1.

The magnitude of the force of +Q or -Q on qo at r = 3 m is FqQ = kQqo/r2 = (9.0 x 109 N-m2/C2)(2 x 10-6 C)(10-12 C)/(9 m2) = 2 x 10-9 N.

The force of +Q on +q0 at

P1 is to the right since +Q repels +qo and urges it to the right

P2 is to the left since +Q repels +qo and urges it to the left.

The force of -Q on + qo at

P1 is to the left since -Q attracts +qo and urges it to the left

P2 to the right since -Q attracts +qo and urges it to the right.

2.

By definition, the direction of the electric field is the direction in which a positive test charge is urged. The magnitude of the electric field at a point P is equal to the electric force Fe on a test charge qo divided by the test charge.

For all points in Fig. 1 above, the magnitude of the electric field = (Fe)/qo = (kQqo/r2)/qo = kQ/r2 = (9 x 109 N-m2/C2)(2 x 10-6 C)/9 m2 = 2 x 103 N/C.

In Fig. 1a, the electric field at P1 is to the right because a positive test charge there would be urged to the right. At P2 the electric field is to the left because a positive test charge there would be urged to the left.

In Fig. 1b, the electric field at P1 is to the left because a positive test charge there would be urged to the left. At P2 the electric field is to the right because a positive test charge there would be urged to the right. Do not use the sign of a charge setting up a field to determine the direction of a field. Go to the point, imagine placing a positive test charge there, and ask what direction the test charge would be urged by the charge setting up the field.

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