
#1
Jan1611, 09:26 PM

P: 3

A proton is in a vacuum near the surface of Earth. Where should a second proton be placed so that the electrostatic force it exerts on the first proton balances the weight of the first proton?
F = (k×Abs[q1]×Abs[q2]) / r^2 k = 8.99×10^9 Nm^2/C^2 



#2
Jan1611, 09:51 PM

HW Helper
P: 6,210

You need to post your attempt.




#3
Jan1611, 10:31 PM

P: 3

The first attempt, I got 1.51911E14 which was incorrect.
It says "Did you use the Coulomb equation to get the electrostatic force? Do you recall how to calculate the gravitational force on an object (the weight) near Earth's surface?" I did do all of the above, but with no success. proton = 1.602176 10^19; Solve[1 == (k proton^2)/r^2, r] {{r > 1.51911*10^14}, {r > 1.51911*10^14}} This was the answer that was incorrect. I thought that the left side of the equation was 1, for some reason, I'm not sure. If it is 0, then the equation is unsolvable. What did I do wrong? 



#4
Jan1611, 10:38 PM

HW Helper
P: 6,210

Electrostatic Force Exerted Between Two ProtonsYou want the weight to be the same the electric force. So you need to equate mg to the formula for the electrostatic force. 



#5
Jan1811, 02:54 PM

P: 3

I did 1.6762622*10^27*9.80665 == 8.99*10^9*1.602*10^19*1.602*10^19/r^2
and solved for r and got the right answer! Thank you! Solve[1.6762622*10^27*9.80665 == 8.99*10^9*1.602*10^19*1.602*10^19/r^2, r] {{r > 0.118471}, {r > 0.118471}} r=0.118471m 


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