Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
@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?? "
 one year ago
 one year 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?? "
 one year ago
 one year ago

This Question is Closed

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

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

MashyBest ResponseYou've already chosen the best response.0
there you go!!.. so whats the problem ?
 one year ago

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

UnkleRhaukusBest ResponseYou've already chosen the best response.3
notice how the distance cancels out,
 one year ago

MashyBest ResponseYou've already chosen the best response.0
oh lol you opened this question to explain it to someone lol.. sorry :D
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.3
can close the question now @azooo ?
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.