First slide

Redistribution of charges and concept of common potential

Question

Two insulated charged conducting spheres of radii 20 cm and 15 cm respectively and having an equal charge of 10C are connected by a copper wire and then they are separated. Then

Easy

A

Both the spheres will have the same charge of 10 C
B

Surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere
C

Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere
D

Surface charge density on the two spheres will be equal

Solution

After redistribution, charges on them will be different, but they will acquire common potential
$i.e. k \frac{Q_{1}}{r_{1}} = k \frac{Q_{2}}{r_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}$
As $σ = \frac{Q}{4 {πr}^{2}} \Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}} \Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow σ \propto \frac{1}{r}$
i.e. surface charge density on smaller sphere will be more.

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