Question Solved1 Answer An electron at Earth's surface experiences a gravitational force meg. How far away can a proton be and still produce the same force on the electron? (Your answer should show why gravity is unimportant on the molecular scale!)

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An electron at Earth's surface experiences a gravitational force meg. How far away can a proton be and still produce the same force on the electron? (Your answer should show why gravity is unimportant on the molecular scale!)

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General guidance Concepts and reasonThe concept required to solve the given problem is Coulomb’s law and gravitational force.Initially, calculate the gravitational force that acts on the electron.Compute the electrostatic force between electron and proton with the help of Coulomb’s law and equate it to the gravitational force to compute the expression for separation between the electron and proton.Finally, substitute the corresponding values into the expression thus obtained to compute the value of separation between electron and proton. FundamentalsCoulomb’s Law: The law states that the electrostatic force between any two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.It is given by, FAB=keqAqBrBA2{F_{{\rm{AB}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{A}}}{q_{\rm{B}}}}}{{{r_{{\rm{BA}}}}^2}}FAB​=rBA​2ke​qA​qB​​ Here, ke{k_e}ke​ is the coulomb’s constant, qAandqB{q_{\rm{A}}}{\rm{ and }}{q_{\rm{B}}}qA​andqB​ are the charges and rBA{r_{{\rm{BA}}}}rBA​ is the distance between two charges qAandqB{q_{\rm{A}}}{\rm{ and }}{q_{\rm{B}}}qA​andqB​ . Gravitational force: It is the force with which earth attracts object towards itself. It is given by the formula, F=mgF = mgF=mg Here, mmm is the mass and ggg is the acceleration due to gravity. Show more First Step | All Steps | Answer Only Step-by-step Step 1 of 2 The gravitational force experienced by the electron will be given by, Fe=meg{F_{\rm{e}}} = {m_{\rm{e}}}gFe​=me​g Here, me{m_{\rm{e}}}me​ is the mass of electron and ggg is the acceleration due to gravity.Substitute 9.1×10−31kg9.1 \times {10^{ - 31}}{\rm{ kg}}9.1×10−31kg for me{m_e}me​ and 9.8m/s29.8{\rm{ m / }}{{\rm{s}}^2}9.8m/s2 for ggg in the above equation. Fe=(9.1×10−31kg)(9.8m/s2)=89.2×10−31N\begin{array}{c}\\{F_{\rm{e}}} = \left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)\left( {9.8{\rm{ m / }}{{\rm{s}}^2}} \right)\\\\ = 89.2 \times {10^{ - 31}}{\rm{ N}}\\\end{array}Fe​=(9.1×10−31kg)(9.8m/s2)=89.2×10−31N​ Explanation | Common mistakes | Hint for next stepThe gravitational force acting on the electron was calculated by the formula, F=mgF = mgF=mg . Step 2 of 2 The electrostatic force that acts between the electron and the proton will be, F=keq1q2r2F = \frac{{{k_{\rm{e}}}{q_1}{q_2}}}{{{r^2}}}F=r2ke ... See the full answer