How Do Three Connected Conducting Spheres Share Total Charge?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
mitleid
Messages
56
Reaction score
1
Was curious how some of you guys would solve this problem...

Three conducting spheres of radii a, b and c are connected by negligibly thin conducting wires. Distances between the spheres are much larger than their sizes. The electric field on the surface of a is measured to be E[tex]_{a}[/tex]. What is the total charge Q that this system of three spheres holds?

E = Q/r[tex]^{2}[/tex]*Ke

Q = Q[tex]_{a}[/tex]+Q[tex]_{b}[/tex]+Q[tex]_{c}[/tex]

The way I solved it is most likely not the way my professor intended. I said that since the amount of charge on each sphere is a function only of the radius of the sphere...

a + b + c = x

a/x = percentage of Q shared on sphere a (called this S[tex]_{a}[/tex])

so Q[tex]_{a}[/tex] = S[tex]_{a}[/tex]*Q
and Q = Q[tex]_{a}[/tex]/S[tex]_{a}[/tex]

I imagine I'm missing a conceptual link that'd make another path to solving this more clear.
 
Physics news on Phys.org
The charge surface density would equal, so the total Q is distributed according to the fraction of surface area.

Determine the surface area for each sphere and total of all three, then ratio the area of each sphere to the total.

Area of sphere is proprotional to r2, where r is the radius.