Recent content by hbc.football2

I got that the electric field should equal 8.1877E-15 for Li+7 because 1.62E-14-8.1877E-15= centripetal force 8.0123E-15

for my final answer I got that the potential should be lowered by 2.193E-17 V for the Li+6 to travel the same path.

So I think I am seeing the end of the problem I can plug in the E for Li+7 to find lambda which then will help me find delta V then plug in the E for Li+6 to find lambda which will then find that particular delta V which means that I will need to reduce my potential for the Li+6 so my answer...

would my r be .003 meters because that's the distance from the radius of the traveling particle?

Can I use E to find lambda? E=lambda/2(pi)(epsilon-nought)r? and then plug it into the equation to find delta V? How else would I find the charge density of the capacitor plates?

So now that I have E you mentioned finding a formula that has electric potential for the capacitor plates. I looked at all my fomulas and couldn't find a capacitor related equation that would work because of my limited data. Could I use E/q=v?

are you finding electric field strength from F=Eq? Is the q the charge of the ion?

I'm a little confused on how to find the potential between the plates because I am not given any explicit information about the plates themselves. Can I use the integral of E(dot)dl from a to b to find the electric potential?

This is similar except the semicircular chamber is closed flip the picture counterclockwise 90 degrees.

There is no distinction between plate A and plate B.

I found that the Magnetic force for Li+7 is 1.62*10^-14 and for Li+6 is 1.255*10^-14 leaving me with that for Li+7 the electric force needs to equal 8.1877E-15 in the opposite direction of the magnetic force. and for Li+6 4.5375E-15 in the opposite direction of the magnetic force.

For some reason I can't get the picture to show up. I tried multiple ways. The picture is rather simple, it is a rainbow/half donut shape with a path going through which is designated as R (goes from the origin to the path). R1 is the radius of the bottom of the rainbow from the origin and R2...

Would there be a negative Vab for the Li+6 because once the plates become charged more the potential goes down?

Okay, so I found my magnetic force to be: LI+7 3.4*10^-40 Li+6 3.145*10^-37 using the F=mv^/r formula for centripetal force I found that: Li+7: 8.0123*10^-15 and Li+6: 8.0125*10^-15 is needed for a radius of .048. I'm lost. Should I subtract the magnetic force form the centripetal to...

Sorry, I would use Eq=qvB to find E then divide the difference by q from Li+6 to Li+7 to get the potential