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Chapter 21: Problem 28

Another unit of charge is the electrostatic unit (esu). It is defined as follows: Two point charges, each of 1 esu and separated by \(1 \mathrm{~cm}\), exert a force of exactly 1 dyne on each other: 1 dyne \(=1 \mathrm{~g} \mathrm{~cm} / \mathrm{s}^{2}=1 \cdot 10^{-5} \mathrm{~N}\). a) Determine the relationship between the esu and the coulomb. b) Determine the relationship between the esu and the elementarv charoe

Short Answer

Expert verified
Answer: (a) 1 esu is approximately equal to 3.34 x 10^-10 C, and (b) 1 esu is approximately equal to 4.8 x 10^-10 elementary charges.

Step by step solution

01

(Step 1: Recalling the definition of esu)

An electrostatic unit (esu) is defined as a unit of charge such that two point charges, each of 1 esu and separated by 1 cm, exert a force of exactly 1 dyne on each other. Additionally, 1 dyne is equal to \(1\times 10^{-5}\) N.
02

(Step 2: Using Coulomb's law)

Coulomb's law states that the electrostatic force (F) between two charged particles is given by: $$ F = k {\frac{q_1 \cdot q_2}{r^2}} $$ Where \(k\) is Coulomb's constant (\(8.99 \times 10^{9} \dfrac{Nm^2}{C^2}\)), \(q_1\) and \(q_2\) are the respective charges, and \(r\) is the distance between the charges. For our case, the given force (\(F_{esu}\)) is 1 dyne (\(1\times 10^{-5}\) N), \(q_1=q_2=1 \text{ esu}\), and \(r=1 \text{ cm}=0.01 \text{ m}\).
03

(Step 3: Solve for the relationship between esu and coulomb)

Now we plug the values into Coulomb's law formula to find the relationship between esu and coulomb. $$1\times 10^{-5} \mathrm{N} = k \frac{(1 \text{ esu})^2}{(0.01)^2} $$ Solving for esu: $$ 1 \text{ esu} = \sqrt{\frac{1\times 10^{-5} \mathrm{N} \cdot (0.01)^2}{k}} $$ Plug in the value of \(k\) to get the conversion factor between esu and coulomb: $$ 1 \text{ esu} = \sqrt{\frac{1\times 10^{-5} \mathrm{N} \cdot (0.01)^2}{8.99 \times 10^{9} \dfrac{Nm^2}{C^2}}} $$ Calculating this, we get: $$ 1 \text{ esu} \approx 3.34\times 10^{-10} \text{ C}$$
04

(Step 4: Determine the relationship between esu and elementary charge)

We know that the elementary charge (\(e\)) is equal to \(1.6 \times 10^{-19} \text{ C}\). To find the relationship between esu and the elementary charge, we will divide the elementary charge by the conversion factor found in Step 3: $$ \frac{e}{1 \text{ esu}} = \frac{1.6 \times 10^{-19} \text{ C}}{3.34\times 10^{-10} \text{ C}} $$ Calculating this, we get: $$ \frac{e}{1 \text{ esu}} \approx 4.8\times 10^{-10} $$ So, the relationship between the esu and the elementary charge is: $$ 1 \text{ esu} \approx 4.8\times 10^{-10} \text{ elementary charges} $$

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Most popular questions from this chapter

Performing an experiment similar to Millikan's oil drop experiment, a student measures these charge magnitudes: \(3.26 \cdot 10^{-19} \mathrm{C} \quad 5.09 \cdot 10^{-19} \mathrm{C} \quad 1.53 \cdot 10^{-19} \mathrm{C}\) \(6.39 \cdot 10^{-19} \mathrm{C} \quad 4.66 \cdot 10^{-19} \mathrm{C}\) Find the charge on the electron using these measurements.

Eight \(1.00-\mu C\) charges are arrayed along the \(y\) -axis located every \(2.00 \mathrm{~cm}\) starting at \(y=0\) and extending to \(y=14.0 \mathrm{~cm} .\) Find the force on the charge at \(y=4.00 \mathrm{~cm} .\)

A 10.0 -g mass is suspended \(5.00 \mathrm{~cm}\) above a nonconducting flat plate, directly above an embedded charge of \(q\) (in coulombs). If the mass has the same charge, \(q\), how much must \(q\) be so that the mass levitates (just floats, neither rising nor falling)? If the charge \(q\) is produced by adding electrons to the mass, by how much will the mass be changed?

A positive charge \(q_{1}=1.00 \mu \mathrm{C}\) is fixed at the origin, and a second charge \(q_{2}=-2.00 \mu \mathrm{C}\) is fixed at \(x=10.0 \mathrm{~cm} .\) Where along the \(x\) -axis should a third charge be positioned so that it experiences no force?

Two charged objects experience a mutual repulsive force of \(0.10 \mathrm{~N}\). If the charge of one of the objects is reduced by half and the distance separating the objects is doubled, what is the new force?
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