Two metal spheres, one of radius and the other of radius respectively have the same surface charge density . They are brought in contact and separated. What will be the new surface charge densities on them? [2019]

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Solution

Q.
Two metal spheres, one of radius $R$ and the other of radius $2 R$ respectively have the same surface charge density $σ$ . They are brought in contact and separated. What will be the new surface charge densities on them [2019]

1 $σ_{1} = \frac{5}{6} σ, σ_{2} = \frac{5}{2} σ$

2 $σ_{1} = \frac{5}{2} σ, σ_{2} = \frac{5}{6} σ$

3 $σ_{1} = \frac{5}{2} σ, σ_{2} = \frac{5}{3} σ$

4 $σ_{1} = \frac{5}{3} σ, σ_{2} = \frac{5}{6} σ$

Ans.
(4)

Before contact, $Q_{1} = σ \cdot 4 π R^{2} and Q_{2} = σ \cdot 4 π {(2 R)}^{2}$

As, surface charge density, $σ = \frac{Net charge (Q)}{Surface area (A)}$

Now, after contact, $Q_{1}' + Q_{2}' = Q_{1} + Q_{2} = 5 Q_{1}$

$= 5 (σ \cdot 4 π R^{2})$ ...(i)

They will be at equal potentials, so

$\frac{Q_{1}'}{R} = \frac{Q_{2}'}{2 R} \Rightarrow Q_{2}' = 2 Q_{1}'$

$∴$ $3 Q_{1}' = 5 (σ \cdot 4 π R^{2}) (From equation (i))$

$and Q_{2}' = \frac{10}{3} (σ \cdot 4 π R^{2})$

$∴$ $σ_{1} = \frac{5}{3} σ and σ_{2} = \frac{5}{6} σ$

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