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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

Two lightweight metal spheres are suspended near each other from insulating threads. One sphere has a net charge; the other sphere has no net charge. The spheres will a) attract each other. b) exert no net electrostatic force on each other. c) repel each other. d) do any of these things depending on the sign of the charge on the one sphere.

A point charge \(q_{1}=100 . \mathrm{nC}\) is at the origin of an \(x y\) -coordinate system, a point charge \(q_{2}=-80.0 \mathrm{nC}\) is on the \(x\) -axis at \(x=2.00 \mathrm{~m},\) and a point charge \(q_{3}=-60.0 \mathrm{nC}\) is on the \(y\) -axis at \(y=-2.00 \mathrm{~m} .\) Determine the net force (magnitude and direction) on \(q_{1}\).

From collisions with cosmic rays and from the solar wind, the Earth has a net electric charge of approximately \(-6.8 \cdot 10^{5} \mathrm{C} .\) Find the charge that must be given to a \(1.0-\mathrm{g}\) object for it to be electrostatically levitated close to the Earth's surface.

Two equal magnitude negative charges \((-q\) and \(-q)\) are fixed at coordinates \((-d, 0)\) and \((d, 0) .\) A positive charge of the same magnitude, \(q\), and with mass \(m\) is placed at coordinate \((0,0),\) midway between the two negative charges. If the positive charge is moved a distance \(\delta \ll d\) in the positive \(y\) -direction and then released, the resulting motion will be that of a harmonic oscillator-the positive charge will oscillate between coordinates \((0, \delta)\) and \((0,-\delta)\). Find the net force acting on the positive charge when it moves to \((0, \delta)\) and use the binomial expansion \((1+x)^{n} \approx 1+n x,\) for \(x \ll 1,\) to find an expression for the frequency of the resulting oscillation.

As shown in the figure, charge 1 is \(3.94 \mu \mathrm{C}\) and is located at \(x_{1}=-4.7 \mathrm{~m},\) and charge 2 is \(6.14 \mu \mathrm{C}\) and is at \(x_{2}=\) \(12.2 \mathrm{~m} .\) What is the \(x\) -coordinate of the point at which the net force on a point charge of \(0.300 \mu \mathrm{C}\) is zero?
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