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Chapter 16: Problem 9

If the electric force of repulsion between two \(1-\mathrm{C}\) charges is $10 \mathrm{N}$, how far apart are they?

Short Answer

Expert verified
Question: Given that two charges of 1 C each experience an electric force of repulsion of 10 N, find the distance between these two charges. Answer: The distance between the two 1-C charges is approximately \(3.33 \times 10^{-5}\,\mathrm{m}\).

Step by step solution

01

Write down the given values

We are given: - The value of each charge, \(q_1\) and \(q_2\), is \(1\,\mathrm{C}\) - The electric force of repulsion, \(F\), is \(10\,\mathrm{N}\)
02

Write down Coulomb's law formula

Coulomb's law states that the magnitude of the electrostatic force between two charged particles is given by: $$F = k \frac{|q_1 q_2|}{r^2}$$ Where: - \(F\) is the force between the two charges - \(q_1\) and \(q_2\) are the charges - \(r\) is the distance between the two charges - k is the electrostatic constant, which is approximately \(8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}\)
03

Plug in the given values and solve for r

We will now plug in the given values into the formula: $$10\,\mathrm{N} = (8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}) \frac{(1\,\mathrm{C})(1\,\mathrm{C})}{r^2}$$ Divide both sides by \((8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}})(1\,\mathrm{C}^2)\): $$r^2 = \frac{10\,\mathrm{N}}{(8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}})(1\,\mathrm{C}^2)}$$ Now, compute the value of \(r^2\): $$r^2 \approx 1.11 \times 10^{-9}\,\mathrm{m^2}$$
04

Find the value of r

To find the distance between the two charges, take the square root of both sides: $$r = \sqrt{1.11 \times 10^{-9}\,\mathrm{m^2}}$$ Compute the value of r: $$r \approx 3.33 \times 10^{-5}\,\mathrm{m}$$
05

Write down the final answer

The distance between the two \(1-\mathrm{C}\) charges is approximately \(3.33 \times 10^{-5}\,\mathrm{m}\).

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Most popular questions from this chapter

A conducting sphere that carries a total charge of \(+6 \mu \mathrm{C}\) is placed at the center of a conducting spherical shell that also carries a total charge of \(+6 \mu \mathrm{C}\). The conductors are in electrostatic equilibrium. (a) Determine the charge on the inner surface of the shell. (b) Determine the total charge on the outer surface of the shell.
A dipole consists of two equal and opposite point charges \((\pm q)\) separated by a distance \(d .\) (a) Write an expression for the electric field at a point \((0, y)\) on the dipole axis. Specify the direction of the field in all four regions: \(y>\frac{1}{2} d, 0<y<\frac{1}{2} d,-\frac{1}{2} d<y<0,\) and \(y<-\frac{1}{2} d .\) (b) At distant points \((|y| \gg d),\) write a simpler, approximate expression for the field. To what power of \(y\) is the field proportional? Does this conflict with Coulomb's law? [Hint: Use the binomial approximation $\left.(1 \pm x)^{n} \approx 1 \pm n x \text { for } x \ll 1 .\right]$
Suppose a 1.0 -g nugget of pure gold has zero net charge. What would be its net charge after it has \(1.0 \%\) of its electrons removed?
(a) Use Gauss's law to prove that the electric field outside any spherically symmetric charge distribution is the same as if all of the charge were concentrated into a point charge. (b) Now use Gauss's law to prove that the electric field inside a spherically symmetric charge distribution is zero if none of the charge is at a distance from the center less than that of the point where we determine the field.
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
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