Solving Charge in Conductors: Find Q1, Q2 & Combined Potential

[songoku]

Homework Statement

A sphere conductor of radius 18 cm has potential 27 Volt. Another sphere conductor has potential 18 Volt. Both of them are connected and the combined potential is 24 Volt. Find:
a. the charge of second sphere
b. the charge of each sphere now

Homework Equations

Q = CV
V = kQ / r

The Attempt at a Solution

a.
V₁ = k.Q₁ / r₁
27 = 9 x 10⁹ x Q₁ / (18 x 10^-2)
Q₁ = 5.4 x 10^-10 C

Then I don't know how to continue...

 

[AGNuke]

Try using conservation of charge.

EDIT: Are those two sphere of same radius?

 

[ehild]

You can calculate also the final charge Q1' on the first sphere, as its potential is known.
As for the second sphere, kQ₂/R₂=18 and kQ'₂/R₂=24. From here, you can find the ratio Q2'/Q2. As AGNuke said, the sum of the charge on the spheres is conserved: Q1+Q2=Q1'+Q2'.

ehild

 

[songoku]

Try using conservation of charge.

EDIT: Are those two sphere of same radius?

I don't know but maybe they are not

You can calculate also the final charge Q1' on the first sphere, as its potential is known.
As for the second sphere, kQ₂/R₂=18 and kQ'₂/R₂=24. From here, you can find the ratio Q2'/Q2. As AGNuke said, the sum of the charge on the spheres is conserved: Q1+Q2=Q1'+Q2'.

ehild

V' = k Q1' / r₁
24 = 9 x 10⁹ x Q1' / (18 x 10^-2)
Q1' = 4.8 x 10^-10 C

kQ₂/R₂=18 ; kQ'₂/R₂=24
So Q2'/Q2 = 24/18 = 4/3

Q1+Q2=Q1'+Q2'
5.4 x 10^-10 + Q2 = 4.8 x 10^-10 + 4/3 Q2
Q2 = 1.8 x 10^-10 C

Q2' = 2.4 x 10^-10 C


If the spheres have same radius, then the combined potential should be: (27 + 18)/2 = 45/2 V. Am I correct in this case?

Thanks

 

[ehild]

I don't know but maybe they are not

Q2 = 1.8 x 10^-10 C

Q2' = 2.4 x 10^-10 C


If the spheres have same radius, then the combined potential should be: (27 + 18)/2 = 45/2 V. Am I correct in this case?

Thanks

Your solution is excellent and you are right, if the radii were the same the final voltage would be 22.5 V.

ehild

 

[songoku]

OK. Thanks a lot