/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 What is the ratio of the electri... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the ratio of the electric force to the gravitational force between a proton and an electron separated by \(5.3 \times 10^{-11} \mathrm{~m}\) (the radius of a hydrogen atom)?

Short Answer

Expert verified
Answer: The ratio of the electric force to the gravitational force between a proton and an electron separated by \(5.3 \times 10^{-11} \mathrm{m}\) is approximately \(2.26 \times 10^{39}\).

Step by step solution

01

Write down the given values

We are given that the distance between the proton and electron (r) is \(5.3 \times 10^{-11} \mathrm{~m}\).
02

Calculate the electric force

We will use Coulomb's law to calculate the electric force between the proton and electron. Coulomb's law is given by: $$F_e = k_e \frac{q_1q_2}{r^2}$$ Where, \(k_e\) is the electrostatic constant (\(8.99 \times 10^{9} \mathrm{Nm^2C^{-2}}\)), \(q_1\) and \(q_2\) are the charges of the proton and electron, each of which is \(1.602 \times 10^{-19} \mathrm{C}\), and \(-1.602 \times 10^{-19} \mathrm{C}\), respectively, \(r\) is the distance between particles (\(5.3 \times 10^{-11} \mathrm{m}\)). Now, substitute these values in the formula: $$F_e = (8.99 \times 10^9) \frac{(1.602 \times 10^{-19})(-1.602 \times 10^{-19})} {(5.3 \times 10^{-11})^2}$$ The negative sign in the electron charge indicates the opposite direction of the force, but since we are calculating the magnitude, we can ignore the negative sign. After calculating, we get: $$F_e \approx 8.20 \times 10^{-8} \mathrm{N}$$
03

Calculate the gravitational force

We will use the universal law of gravitation to calculate the gravitational force between the proton and electron. The formula is: $$F_g = G \frac{m_1m_2}{r^2}$$ Where, \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{Nm^2kg^{-2}}\)), \(m_1\) and \(m_2\) are the masses of the proton and electron, which are \(1.673 \times 10^{-27} \mathrm{kg}\) and \(9.110 \times 10^{-31} \mathrm{kg}\), respectively, \(r\) is the distance between particles (\(5.3 \times 10^{-11} \mathrm{m}\)). Now, substitute these values in the formula: $$F_g = (6.674 \times 10^{-11}) \frac{(1.673 \times 10^{-27})(9.110 \times 10^{-31})} {(5.3 \times 10^{-11})^2}$$ After calculating, we get: $$F_g \approx 3.63 \times 10^{-47} \mathrm{N}$$
04

Calculate the ratio of electric force to gravitational force

Now, we can find the required ratio by dividing the electric force (\(F_e\)) by the gravitational force (\(F_g\)): $$\frac{F_e}{F_g} = \frac{8.20 \times 10^{-8}}{3.63 \times 10^{-47}}$$ After calculating, we get: $$\frac{F_e}{F_g} \approx 2.26 \times 10^{39}$$ So, the ratio of the electric force to the gravitational force between a proton and an electron separated by \(5.3 \times 10^{-11} \mathrm{~m}\) (the radius of a hydrogen atom) is approximately \(2.26 \times 10^{39}\).

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Most popular questions from this chapter

Find the total positive charge of all the protons in \(1.0 \mathrm{mol}\) of water.
Two tiny objects with equal charges of \(7.00 \mu \mathrm{C}\) are placed at the two lower corners of a square with sides of \(0.300 \mathrm{m},\) as shown. Find the electric field at point \(B\) midway between the upper left and right corners.
A parallel-plate capacitor consists of two flat metal plates of area \(A\) separated by a small distance \(d\). The plates are given equal and opposite net charges \(\pm q\) (a) Sketch the field lines and use your sketch to explain why almost all of the charge is on the inner surfaces of the plates. (b) Use Gauss's law to show that the electric field between the plates and away from the edges is $E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} \cdot(\mathrm{c})$ Does this agree with or contra- dict the result of Problem \(70 ?\) Explain. (d) Use the principle of superposition and the result of Problem 69 to arrive at this same answer. [Hint: The inner surfaces of the two plates are thin, flat sheets of charge.]
(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.
\(\mathrm{A}+2.0\) -nC point charge is \(3.0 \mathrm{cm}\) away from a $-3.0 \mathrm{-nC}$ point charge. (a) What are the magnitude and direction of the electric force acting on the +2.0 -nC charge? (b) What are the magnitude and direction of the electric force acting on the -3.0 -nC charge?
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