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A conducting sphere of radius 10 cm is charged with 10 μC. Another uncharged sphere of radius 20 cm is allowed to touch it for some time. After that if the spheres are separated, then surface density of charges on the spheres will be in the ratio
- a)1 : 1
- b)2 : 1
- c)1 : 2
- d)4 : 1
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A conducting sphere of radius 10 cm is charged with 10 μC. Another un...
Solution:
Given,
Radius of first sphere, r1 = 10 cm
Charge on first sphere, q1 = 10 μC
Radius of second sphere, r2 = 20 cm
Charge on second sphere, q2 = 0 μC (Initially uncharged)
Let the final charge on the second sphere be q2’.
Step 1: Charge on first sphere will distribute itself on both the spheres
Charge on first sphere, q1 = Charge on first sphere after sharing + Charge on second sphere after sharing
q1 = q1’ + q2’
Step 2: Find the charge transferred from first sphere to second sphere
Both the spheres are in contact, so the charge will transfer from the first sphere to the second sphere until they have the same potential.
The potential of the first sphere, V1 = kq1/r1
The potential of the second sphere, V2 = kq2’/r2
As they are in contact, they will have the same potential, so
V1 = V2
kq1/r1 = kq2’/r2
q2’ = (r2/r1)q1
q2’ = (20/10)10 μC = 20 μC
Step 3: Find the surface charge density on the spheres
Surface charge density, σ = q/4πr2
For the first sphere, σ1 = q1/4πr1² = 10/(4π×10²) = 0.0008 μC/m²
For the second sphere, σ2 = q2’/4πr2² = 20/(4π×20²) = 0.0002 μC/m²
Step 4: Find the ratio of surface charge density on the spheres
σ1/σ2 = (0.0008 μC/m²)/(0.0002 μC/m²) = 4
Therefore, the surface density of charges on the spheres will be in the ratio 2:1 (Option B).
Given,
Radius of first sphere, r1 = 10 cm
Charge on first sphere, q1 = 10 μC
Radius of second sphere, r2 = 20 cm
Charge on second sphere, q2 = 0 μC (Initially uncharged)
Let the final charge on the second sphere be q2’.
Step 1: Charge on first sphere will distribute itself on both the spheres
Charge on first sphere, q1 = Charge on first sphere after sharing + Charge on second sphere after sharing
q1 = q1’ + q2’
Step 2: Find the charge transferred from first sphere to second sphere
Both the spheres are in contact, so the charge will transfer from the first sphere to the second sphere until they have the same potential.
The potential of the first sphere, V1 = kq1/r1
The potential of the second sphere, V2 = kq2’/r2
As they are in contact, they will have the same potential, so
V1 = V2
kq1/r1 = kq2’/r2
q2’ = (r2/r1)q1
q2’ = (20/10)10 μC = 20 μC
Step 3: Find the surface charge density on the spheres
Surface charge density, σ = q/4πr2
For the first sphere, σ1 = q1/4πr1² = 10/(4π×10²) = 0.0008 μC/m²
For the second sphere, σ2 = q2’/4πr2² = 20/(4π×20²) = 0.0002 μC/m²
Step 4: Find the ratio of surface charge density on the spheres
σ1/σ2 = (0.0008 μC/m²)/(0.0002 μC/m²) = 4
Therefore, the surface density of charges on the spheres will be in the ratio 2:1 (Option B).
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Community Answer
A conducting sphere of radius 10 cm is charged with 10 μC. Another un...
When the charged sphere in contact with uncharged sphere, the charge will flow till the potential at the surface of the spheres become same.
Let the radius of the charged sphere is r1 and charge density σ1 and charge on it be q1 and the radius of the uncharged sphere is r2 and charge density σ2 and charge on it be q2 after both are in contact with each other.
Charge density 
Hence, 
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A conducting sphere of radius 10 cm is charged with 10 μC. Another uncharged sphere of radius 20 cm is allowed to touch it for some time. After that if the spheres are separated, then surface density of charges on the spheres will be in the ratioa)1 : 1b)2 : 1c)1 : 2d)4 : 1Correct answer is option 'B'. Can you explain this answer?
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