Electromagnetism problem: Merging of 2 charged drops of mercury

AI Thread Summary
The discussion revolves around the merging of two charged mercury drops and the conflicting solutions regarding their potential and radius after merging. Initial calculations suggest that the radii of the drops are R1 = 0.514 and R2 = 0.54, leading to a combined radius R3 of 0.664 based on volume conservation. Charge conservation yields a total charge Q3 of 70 nC, resulting in a potential V of 948V. However, a classmate obtained a different potential of 614V, highlighting discrepancies in the calculations. The conversation also touches on the use of LaTeX for better expression of mathematical equations.
Elj
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Homework Statement
Two non-identical spherical drops of mercury one with charge of 30.0 nC and a potential
of 500.0 V and the other one with charge of 40.0 nC and potential of 700.0 Vat the
surface. The two drops merge to form a single drop. What is the potential at the surface
of the new drop?
Relevant Equations
V=kq/r v=4/3pir^3
I originally thought that this problem was simple, and it still seems like it is, but there are conflicting solutions and I don't know which is correct. So I first solved for R1 and R2 using V=kQ/r where R1 is 0.514 and R2 is 0.54. My original thought was volume is conserved so V1 + V2 = V3 and rearranging to get the radius then R3 = (R2^3 + R1^3)^3 = 0.664. Then the charge is conserved so Q1 + Q2 = Q3 so Q3 = 70nc and then plugging it back in to get the potential V=kQ3/R3, V = 948V, but a classmate of mine seemed to get 614V.

Sorry about the format I don't know how to write in Laytex.
 
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I confirm your result.
 
Elj said:
Sorry about the format I don't know how to write in Laytex.
There's link 'LaTeX guide' at the lower left of the edit window.
Enclosing your expression in double # is already a good start:
Elj said:
Relevant Equations: ##V=kq/r ##
and a backslash turns pi into ##\pi##: v=4/3 \pi r^3 yields
Elj said:
##v=4/3\pi r^3##
Underscore gives subscripts
Elj said:
... so ##V_1 + V_2 = V_3## and rearranging to get the radius then ##R_3 = (R_2^3 + R_1^3)^{1/3} = 0.664.##

Sorry about the format I don't know how to write in Laytex.
and in a few minutes you're no longer a Layman...:smile:

Learning goes fast if you right-click ##\TeX## to show the input

and it's fun...

##\ ##
 
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