Chapter 14: Q12Q (page 578)

Figure 14.69 shows a neutral, solid piece of metal placed near two points charges. Copy this diagram.
(a) On your diagram, show the polarization of the piece of metal.
(b) Then, at location A inside the solid piece of metal, carefully draw and label three vectors: (1) ${\vec{E}}_{1}$ , the electric field due to $- q_{1}$ ; (2) ${\vec{E}}_{2}$ , the electric field due to $+ q_{2}$ ; (3) ${\vec{E}}_{3}$ , the electric field due to all of the charges on the metal.
(c) Explain briefly why you drew the vectors the way you did.

Short Answer

Expert verified

In the diagram of the charges present, polarize the metal piece. The positive charge attracts the metal's free electrons, leaving an electron deficiency or, to put it another way, a positive charge attracting the negative charge.
The vector fields ${\vec{E}}_{1}$ were drawn, due to the charge $- q_{1}$ , the vector field ${\vec{E}}_{2}$ due to the charge $+ q_{2}$ and the resulting vector field ${\vec{E}}_{3}$ .
The positive charges produce a force and hence an electric field ${\vec{E}}_{2}$ in the opposite direction of the charge-point X line. At point A, the negative charge causes an attractive force towards it, and its electric field ${\vec{E}}_{1}$ follows the force. The sum of these two vectors produces the final field ${\vec{E}}_{3}$ .

Step by step solution

Identification of given data

The given data can be listed below,

The electric field ${\vec{E}}_{1}$ due to charge $- q_{1}$ .
The electric field ${\vec{E}}_{2}$ due to charge $+ q_{2}$ .

Concept/Significance of polarization

The term polarization, also known as electric polarization, is the alignment of the dipole moments of permanent or induced dipoles in the direction of an applied electric field.

(a) Determination of the polarization of the piece of metal in a diagram.

The polarization of charges is shown below in the diagram.

Thus, in the diagram of the charges present, polarize the metal piece. The positive charge attracts the metal's free electrons, leaving an electron deficiency or, to put it another way, a positive charge attracting the negative charge.

(b) Determination of the electric field due to all of the charges on the metal

The diagram for electric fields is given below,

Thus, the vector fields ${\vec{E}}_{1}$ were drawn, due to the $- q_{1}$ charge, the vector field ${\vec{E}}_{2}$ due to the charge and the resulting vector field ${\vec{E}}_{3}$ .

(c) Explanation of briefly why you drew the vectors the way you did

Positive charge generates a force and, as a result, an electric field ${\vec{E}}_{2}$ in the opposite direction of the charge-point X line. The negative charge at point A attracts it with its electric field ${\vec{E}}_{1}$ , which is parallel to the force. The resulting field ${\vec{E}}_{1}$ is obtained by summing these two vectors.

Thus, the positive charges produce a force and hence an electric field ${\vec{E}}_{2}$ in the opposite direction of the charge-point X line. At point A, the negative charge causes an attractive force towards it, and its electric field ${\vec{E}}_{1}$ follows the force. The sum of these two vectors produces the final field ${\vec{E}}_{3} .$