Electromagnetism problem: Merging of 2 charged drops of mercury

[Elj]

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.

 

[haruspex]

I confirm your result.

 

[BvU]

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:

Relevant Equations: ##V=kq/r ##

and a backslash turns pi into ##\pi##: v=4/3 \pi r^3 yields

##v=4/3\pi r^3##

Underscore gives subscripts

... 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...

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

and it's fun...

##\ ##