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A conducting sphere of radius 10 cm is charged with 10 μC. Another uncharged sphere of radius 20 cm in allowed to touch it for some time. If both the spheres are separated, then surface density of charges on the spheres will be in the ratio of
- a)1:4
- b)4:1
- c)1:2
- d)2:!
Correct answer is option 'D'. Can you explain this answer?
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A conducting sphere of radius 10 cm is charged with 10 μC. Another unc...
Given:
Radius of first sphere, r1 = 10 cm
Charge on first sphere, Q1 = 10 μC
Radius of second sphere, r2 = 20 cm
To find:
Surface density of charges on the spheres after they are separated
Solution:
When the uncharged sphere touches the charged sphere, some of the charge is transferred from the charged sphere to the uncharged sphere. The charge on the two spheres after they are separated can be found using the principle of conservation of charge.
Charge on first sphere before touching, Q1 = 10 μC
Charge on second sphere before touching, Q2 = 0 μC
Total charge before touching, QT = Q1 + Q2 = 10 μC
Let q be the charge transferred from the first sphere to the second sphere
Charge on first sphere after touching, Q1' = Q1 - q
Charge on second sphere after touching, Q2' = Q2 + q
Total charge after touching, QT' = Q1' + Q2' = Q1 + Q2 = 10 μC
Using the principle of conservation of charge, we have:
Q1 - q + q = Q2 + q - q
Q1 = Q2
This means that the two spheres have the same charge after they are separated.
Surface density of charges on a sphere is defined as the charge per unit area of the sphere's surface.
Surface area of first sphere, A1 = 4πr1^2
Surface area of second sphere, A2 = 4πr2^2
Surface density of charges on first sphere, σ1 = Q1 / A1
Surface density of charges on second sphere, σ2 = Q2 / A2
Since Q1 = Q2, we have:
σ1 / σ2 = Q1 / A1 / Q2 / A2
σ1 / σ2 = A2 / A1
σ1 / σ2 = (4πr2^2) / (4πr1^2)
σ1 / σ2 = r2^2 / r1^2
σ1 / σ2 = (20 cm / 10 cm)^2
σ1 / σ2 = 4
Therefore, the surface density of charges on the spheres will be in the ratio of 4:1, which is option D.
Radius of first sphere, r1 = 10 cm
Charge on first sphere, Q1 = 10 μC
Radius of second sphere, r2 = 20 cm
To find:
Surface density of charges on the spheres after they are separated
Solution:
When the uncharged sphere touches the charged sphere, some of the charge is transferred from the charged sphere to the uncharged sphere. The charge on the two spheres after they are separated can be found using the principle of conservation of charge.
Charge on first sphere before touching, Q1 = 10 μC
Charge on second sphere before touching, Q2 = 0 μC
Total charge before touching, QT = Q1 + Q2 = 10 μC
Let q be the charge transferred from the first sphere to the second sphere
Charge on first sphere after touching, Q1' = Q1 - q
Charge on second sphere after touching, Q2' = Q2 + q
Total charge after touching, QT' = Q1' + Q2' = Q1 + Q2 = 10 μC
Using the principle of conservation of charge, we have:
Q1 - q + q = Q2 + q - q
Q1 = Q2
This means that the two spheres have the same charge after they are separated.
Surface density of charges on a sphere is defined as the charge per unit area of the sphere's surface.
Surface area of first sphere, A1 = 4πr1^2
Surface area of second sphere, A2 = 4πr2^2
Surface density of charges on first sphere, σ1 = Q1 / A1
Surface density of charges on second sphere, σ2 = Q2 / A2
Since Q1 = Q2, we have:
σ1 / σ2 = Q1 / A1 / Q2 / A2
σ1 / σ2 = A2 / A1
σ1 / σ2 = (4πr2^2) / (4πr1^2)
σ1 / σ2 = r2^2 / r1^2
σ1 / σ2 = (20 cm / 10 cm)^2
σ1 / σ2 = 4
Therefore, the surface density of charges on the spheres will be in the ratio of 4:1, which is option D.
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A conducting sphere of radius 10 cm is charged with 10 μC. Another uncharged sphere of radius 20 cm in allowed to touch it for some time. If both the spheres are separated, then surface density of charges on the spheres will be in the ratio ofa)1:4b)4:1c)1:2d)2:!Correct answer is option 'D'. Can you explain this answer?
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