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An electric field of magnitude 190  N/C is applied to a solution containing chloride ions. The mobility of chloride ions in solution is 7.91×10-8  (m/s)/(N/C).What is the average drift speed of the chloride ions in the solution?

Short Answer

Expert verified

1.5×10-5m/s

Step by step solution

01

Identification of given data

  • The magnitude of the electric field is, E=190N/C
  • The mobility of chloride ions is, μ=7.91×10-8 m/s/N/C
02

Explanation of Drift speed

The drift speed can be determined by taking the product of the electric field and the mobility of ions. It is the average velocity attained by the charged particles in the presence of the electric field.

03

Determination of the average drift speed of chloride ions

The expression for the drift speed of ions is as follows,

ν=E×μ

Here,Eis magnitude of the electric field, and μis the mobility of chloride ions.

For E=190N/Cand μ=7.91×10-8 m/s/N/C.

ν=190N/C7.91×10-8m/s/N/C=1.5×10-5m/s

Thus, the average drift speed of chloride ions in the solution is 1.5×10-5m/s.

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Most popular questions from this chapter

An electric field is applied to a solution containing bromide ions. As a result, the ions move through the solution with an average drift speed of 3.7×10-7m/s. The mobility of bromide ions in solution is 8.1×10-8(m/s)(N/C). What is the magnitude of the net electric field inside the solution?

: A thin, hollow spherical plastic shell of radius \({\bf{R}}\)carries a uniformly distributed negative charge \({\bf{ - Q}}\). A slice through the plastic shell is shown in Figure 14.95. To the left of the spherical shell are four charges packed closely together as shown (the distance \({\bf{s}}\) is shown greatly enlarged for clarity). The distance from the center of the four charges to the center of the plastic shell is \({\bf{L}}\) , which is much larger than \({\bf{s}}\left( {{\bf{L}} \gg {\bf{s}}} \right)\). Remember that a uniformly charged sphere makes an electric field as though all the charge were concentrated at the center of the sphere.

(a)Calculate the \({\bf{x}}\) and \({\bf{y}}\) components of the electric field at location B, a distance \({\bf{b}}\) to the right of the outer surface of the plastic shell. Explain briefly, including showing the electric field on a diagram. Your results should not contain any symbols other than the given quantities \({\bf{R,Q,q,s,L}}\), and \({\bf{b}}\)(and fundamental constants). You need not simplify the final algebraic results except for taking into account the fact that \({\bf{L}} \gg {\bf{s}}\).

(b)What simplifying assumption did you have to make in part (a)?

(c)The plastic shell is removed and replaced by an uncharged metal ball, as in Figure 14.96. At location Ainside the metal ball, a distance \({\bf{b}}\)to the left of the outer surface of the ball, accurately draw and label the electric field\({{\bf{\vec E}}_{{\bf{ball}}}}\) due to the ball charges and the electric field \({{\bf{\vec E}}_{\bf{4}}}\) of the four charges. Explain briefly.

(d)Show the distribution of ball charges.

(e)Calculate the \({\bf{x}}\) and \({\bf{y}}\) components of the net electric field at location A.

(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)?

In Figure 14.84 there is a permanent dipole on the left with dipole moment μ1=Qs1 and a neutral atom on the right with polarizabilityα , so that it becomes an induced dipole with dipole moment μ2=Qs2=αE1, whereE1 is the magnitude of the electric field produced by the permanent dipole. Show that the force the permanent dipole exerts on the neutral atom isF≈(14πε0)212αμ12r7

You have two identical neutral metal spheres labeled A and B, mounted on insulating posts, and you have a plastic pen that charges negatively when you rub it on your hair (Figure 14.77).


(a) (+ and −) Explain in detail, including diagrams, what operations you would carry out to give sphere A some positive charge and sphere B an equal amount of negative charge. (b) (+ and +) Explain in detail, including diagrams, what operations you would carry out on the neutral spheres to give sphere A some positive charge and sphere B an equal amount of positive charge (the spheres are initially uncharged).

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