Question

Select the correct answer
There are two spheres, placed far apart. The ratio of their radii is 10:1. A charge Q is place
the bigger sphere. They are now connected by a long conducting wire. After this connecti
what would be the ratio of the charge on bigger and smaller spheres?
o 10:1
01:2
o 1:10
02:1
Finish C
Clear Response
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Answer 1

Answer:

10:1

Explanation:

two spheres placed far apart

Answer 2

Answer: (A) 10:1

Explanation:

• Spheres are connected to each other, which means the potential on both spheres are equal
• V= kQ/r
• Charge on the bigger sphere (given) = Q
• Charge on the smaller sphere after the spheres are connected = x
• Charge on the bigger sphere after the spheres are connected = Q-x

∴ V(bigger sphere)=V(smaller sphere) ⇒ $\frac{k(Q-x)}{r_{1} }$ =$\frac{k(x)}{r_{2} }$

Where $\frac{r_{1}}{r_2}$= 10:1 (given)

So, upon substituting and solving,

10x=Q-x ⇒ x=Q/11

Charge on bigger sphere (final) = Q-x= Q- Q/11 = 10Q/11

∴ Ratio of charges between the bigger and smaller sphere = 10Q/11 ÷ Q/11 =

  10:1