Question

(a)The positively charged particle shown in diagram 1 in Figure 14.94 creates an electric field \({{\bf{\vec E}}_{\bf{p}}}\) at location A. Which of the arrows (a–j) in Figure 14.94 best indicates the direction of \({{\bf{\vec E}}_{\bf{p}}}\) at location A?

(b)Now a block of metal is placed in the location shown in diagram 2 in Figure 14.94. Which of the arrows (a–j) in Figure 14.94 best indicates the direction of the electric field \({{\bf{\vec E}}_{\bf{m}}}\) at location Adue only to the charges in and/or on the metal block?

(c)\(\left| {{{{\bf{\vec E}}}_{\bf{p}}}} \right|\)is greater than \(\left| {{{{\bf{\vec E}}}_{\bf{m}}}} \right|\). With the metal block still in place, which of the arrows (a–j) in Figure 14.94 best indicates the direction of the net electric field at location A?

(d)With the metal block still in place, which of the following statements about the magnitude of \({{\bf{\vec E}}_{\bf{p}}}\), the field due only to the charged particle, is correct?

(1) \(\left| {{{{\bf{\vec E}}}_{\bf{p}}}} \right|\)is less than it was originally, because the block is in the way.

(2) \(\left| {{{{\bf{\vec E}}}_{\bf{p}}}} \right|\)is the same as it was originally, without the block.

(3) \(\left| {{{{\bf{\vec E}}}_{\bf{p}}}} \right|\)is zero, because the electric field due to the particle can’t go through the block.

(e)With the metal block still in place, how does the magnitude of\({{\bf{\vec E}}_{{\bf{net}}}}\) at location Acompare to the magnitude of \({{\bf{\vec E}}_{\bf{p}}}\)?

(f)Which of the arrows (a–j) in Figure 14.94 best indicates the direction of the net electric field at the center of the metal block (inside the metal)?

Short answer

(a) The direction of \({{\bf{\vec E}}_{\bf{p}}}\)at location A is best indicated by arrow'c'.

(b) The direction of\({{\bf{\vec E}}_{\bf{m}}}\)at location A is best indicated by arrow'c'.

(c) The direction of the net electric field at location A is best indicated by arrow'c'.

(d) (2)\(\left| {{{{\bf{\vec E}}}_{\bf{p}}}} \right|\)is the same as it was originally, without the block.

(e) Magnitude of\({{\bf{\vec E}}_{{\bf{net}}}}\) at location Ais greater than the magnitude of\({{\bf{\vec E}}_{\bf{p}}}\).

(f) The direction of the net electric field at the center of the metal block is best indicated by 'j'.

Step-by-step solution

Given data

A positively charged particle produces an electric field \({\vec E_p}\) at a location A. A block of conductor is placed in between the positively charged particle and location A. The induced charges in the conductor produces electric field \({\vec E_m}\) at location A.

Electric field direction and magnitude

Electric field is directed away from a positive charge and directed towards a negative charge.

Magnitude of electric field decreases with distance from the source charges.

Charges are distributed in a conductor in the presence of an external field such that the net electric field inside the conductor is zero.

Determining the direction of electric field at A due to the positive charge

Since electric field directs away from a positive charge, the field at A due the positive charge to its left \({\vec E_p}\) will be directed towards the right. Hence the arrow 'c' correctly depicts this field.

Determining the direction of electric field at A due to induced charges in the conductor

The electric field due to the positive charge induces negative charges at the left end of the block and positive charges at the right end of the block. Both these induced charges will create electric fields at A. The field due to the induced positive charges points away from the block and the field due to the negative charges points toward the block. But the negative charges are induced at the left of the block and thus are farther away from point A than the positive charges. Thus the field due to the induced negative charges is weaker than the field due to induced positive charges. Hence the net field \({\vec E_m}\) points away from the block. This is depicted by arrow 'c'.

Determining the direction of the net electric field at A

Since both \({\vec E_p}\) and \({\vec E_m}\) points along arrow 'c', the net field at A which is a vector sum of the two fields, is also directed along arrow 'c'.

Determining the change in magnitude of the field at A due to the positive charge in the presence of the conductor

Presence of other charges don't change the magnitude of the field present in the absence of the charges. The net field is the vector sum of the existing field and the new field due to the introduced charges. Thus,\(\left| {{{\vec E}_p}} \right|\)is the same as it was originally, without the block.

Comparing the net field at A in the presence of both the positive charge and block with the field at A in the absence of the block

As discussed above, both \({\vec E_p}\) and \({\vec E_m}\) are directed to the right. Thus the magnitude of the net field at A \({\vec E_{{\rm{net}}}}\) will be the scalar sum of the magnitudes of \({\vec E_p}\) and \({\vec E_m}\). Hence magnitude of \({\vec E_{{\rm{net}}}}\) is greater than magnitude of \({\vec E_p}\).

Determining the direction of the net electric field at the center of the block

Charges are induced in the block in the presence of the electric field due to the positive charge in such a way that the net field inside the block is zero. This is best depicted by 'j', that is zero magnitude.