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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

An electron beam in an oscilloscope is deflected by the electric field produced by oppositely charged metal plates. If the electric field between the plates is \(2.00 \times 10^{5} \mathrm{N} / \mathrm{C}\) directed downward, what is the force on each electron when it passes between the plates?
Suppose a charge \(q\) is placed at point \(x=0, y=0 .\) A second charge \(q\) is placed at point \(x=8.0 \mathrm{m}, y=0 .\) What charge must be placed at the point \(x=4.0 \mathrm{m}, y=0\) in order that the field at the point $x=4.0 \mathrm{m}, y=3.0 \mathrm{m} \mathrm{be}$ zero?
A flat conducting sheet of area \(A\) has a charge \(q\) on each surface. (a) What is the electric field inside the sheet? (b) Use Gauss's law to show that the electric field just outside the sheet is \(E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} .\) (c) Does this contradict the result of Problem \(69 ?\) Compare the field line diagrams for the two situations.
Two point charges, \(q_{1}=+20.0 \mathrm{nC}\) and \(q_{2}=+10.0 \mathrm{nC},\) are located on the \(x\) -axis at \(x=0\) and \(x=1.00 \mathrm{m},\) respectively. Where on the \(x\) -axis is the electric field equal to zero?
A balloon, initially neutral, is rubbed with fur until it acquires a net charge of \(-0.60 \mathrm{nC} .\) (a) Assuming that only electrons are transferred, were electrons removed from the balloon or added to it? (b) How many electrons were transferred?
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