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@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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which are the formulae you know to calculate the force of gravity between two point masses? and electric force between 2 point charges? :P
\[\begin{align*}\left|\frac{\vec{\mathbf F}_{\text{electric}}}{\vec{\mathbf{F}}_{\text{gravitational}}}\right| &=\frac{K\dfrac{q_{e^-}q_{e^-}}{d^2}}{G\dfrac{m_{e^-}m_{e^-}}{d^2}} % =\frac KG\left( \frac{q_{p}}{m_{p}}\right)^2 \\ %&=\frac{9.0\times10^9 \left[\frac{\text N\cdot\text m^2}{\text C^2}\right]} %{6.7\times10^{-11}\left[\frac{\text{N}\cdot\text m^2}{\text {kg}^2}\right]} %\left(\frac{1.6\times10^{-19}[\text C]}{1.7\times10^{-27}[\text{kg}]}\right)^2 \\ %&=\left(\frac{9.0\times1.6^2}{6.7\times1.7^2}\right)\frac{10^{9-2\times19}}{10^{-11-2\times27}} \\ %&=\left(\frac{9.0\times2.56}{6.7\times2.89}\right)\frac{10^{-9-2\times19}}{10^{-11-54}} \\ %&=1.2\times10^{-29+65} \\ %&=1.2\times10^{36} \end{align*} \]
there you go!!.. so whats the problem ?

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for the rato of the forces between Protons \[\begin{align*} \left|\frac{\vec{\mathbf F}_{\text{electric}}}{\vec{\mathbf{F}}_{\text{gravitational}}}\right| &=\frac{K\dfrac{q_{p}q_{p}}{d^2}}{G\dfrac{m_{p}m_{p}}{d^2}} =\frac KG\left( \frac{q_{p}}{m_{p}}\right)^2 \\ &=\frac{9.0\times10^9 \left[\frac{\text N\cdot\text m^2}{\text C^2}\right]} {6.7\times10^{-11}\left[\frac{\text{N}\cdot\text m^2}{\text {kg}^2}\right]} \left(\frac{1.6\times10^{-19}[\text C]}{1.7\times10^{-27}[\text{kg}]}\right)^2 \\ %&=\left(\frac{9.0\times1.6^2}{6.7\times1.7^2}\right)\frac{10^{9-2\times19}}{10^{-11-2\times27}} \\ &=\left(\frac{9.0\times2.56}{6.7\times2.89}\right)\frac{10^{-9-2\times19}}{10^{-11-54}} \\ %&=1.2\times10^{-29+65} \\ &=1.2\times10^{36} \end{align*}\]
notice how the distance cancels out,
thank u
oh lol you opened this question to explain it to someone lol.. sorry :D
can close the question now @azooo ?
yes thank u

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