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UnkleRhaukus
 3 years ago
@azooo "Two electrons are separated by 1 meter. What is the ratio of the magnitude of the electric force on one electron and the gravitational force on that electron due to the other electron? The charge of an electron is 1.6*10^19C and its mass is 9.1*10^31??
which formuA will use for it?? "
UnkleRhaukus
 3 years ago
@azooo "Two electrons are separated by 1 meter. What is the ratio of the magnitude of the electric force on one electron and the gravitational force on that electron due to the other electron? The charge of an electron is 1.6*10^19C and its mass is 9.1*10^31?? which formuA will use for it?? "

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which are the formulae you know to calculate the force of gravity between two point masses? and electric force between 2 point charges? :P

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.3\[\begin{align*}\left\frac{\vec{\mathbf F}_{\text{electric}}}{\vec{\mathbf{F}}_{\text{gravitational}}}\right &=\frac{K\dfrac{q_{e^}q_{e^}}{d^2}}{G\dfrac{m_{e^}m_{e^}}{d^2}} % =\frac KG\left( \frac{q_{p}}{m_{p}}\right)^2 \\ %&=\frac{9.0\times10^9 \left[\frac{\text N\cdot\text m^2}{\text C^2}\right]} %{6.7\times10^{11}\left[\frac{\text{N}\cdot\text m^2}{\text {kg}^2}\right]} %\left(\frac{1.6\times10^{19}[\text C]}{1.7\times10^{27}[\text{kg}]}\right)^2 \\ %&=\left(\frac{9.0\times1.6^2}{6.7\times1.7^2}\right)\frac{10^{92\times19}}{10^{112\times27}} \\ %&=\left(\frac{9.0\times2.56}{6.7\times2.89}\right)\frac{10^{92\times19}}{10^{1154}} \\ %&=1.2\times10^{29+65} \\ %&=1.2\times10^{36} \end{align*} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there you go!!.. so whats the problem ?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.3for the rato of the forces between Protons \[\begin{align*} \left\frac{\vec{\mathbf F}_{\text{electric}}}{\vec{\mathbf{F}}_{\text{gravitational}}}\right &=\frac{K\dfrac{q_{p}q_{p}}{d^2}}{G\dfrac{m_{p}m_{p}}{d^2}} =\frac KG\left( \frac{q_{p}}{m_{p}}\right)^2 \\ &=\frac{9.0\times10^9 \left[\frac{\text N\cdot\text m^2}{\text C^2}\right]} {6.7\times10^{11}\left[\frac{\text{N}\cdot\text m^2}{\text {kg}^2}\right]} \left(\frac{1.6\times10^{19}[\text C]}{1.7\times10^{27}[\text{kg}]}\right)^2 \\ %&=\left(\frac{9.0\times1.6^2}{6.7\times1.7^2}\right)\frac{10^{92\times19}}{10^{112\times27}} \\ &=\left(\frac{9.0\times2.56}{6.7\times2.89}\right)\frac{10^{92\times19}}{10^{1154}} \\ %&=1.2\times10^{29+65} \\ &=1.2\times10^{36} \end{align*}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.3notice how the distance cancels out,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh lol you opened this question to explain it to someone lol.. sorry :D

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.3can close the question now @azooo ?
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