Question

A conducting sphere of radius 5 cm is charged to 15 μC. Another uncharged

sphere of radius 10 cm is allowed to touch it for enough time. After the two are separated, the surface density of charge on the two spheres will be in what ratio?​

Answer 1

Answer:-

The surface density of charge of the sphere in ratio will be 2:1

• Given:-

Radius of sphere = 5 cm

Charge = 15 μC

Radius of uncharged sphere = 10 cm

• Solution:-

$\red{\bigstar}$ When two conducting spheres touch each other there will be a flow of charge between them until they both have the same potential.

Let R₁ and R₂ be the radius of both sphere.

and

Let Q₁ and Q₂ be the charge on both sphere.

The common potential = V

• The charge on 1st sphere:-

$Q_{1}=4\pi\epsilon_{0}R_{1}$

• The charge on 2nd sphere:-

$Q_{2}=4\pi\epsilon_{0}R_{2}$

• The ratio of the charge of spheres,

$\dfrac{Q_{1}}{Q_{2}}=\dfrac{R_{1}}{R_{2}} \: \: \: \: \: \: \longrightarrow\bf[eqn.1]$

Now,

• The charge density on the 1st sphere

$\sigma_{1}=\dfrac{Q_{1}}{4\pi R_{1}^2}$

• The charge density on the 2nd sphere

$\sigma_{2}=\dfrac{Q_{2}}{4\pi R_{2}^2}$

• The ratio of the charge density

$\dfrac{\sigma_{1}}{\sigma_{2}}=\dfrac{Q_{1}R_{2}^2}{Q_{2}R_{1}^2} \: \: \: \: \: \: \longrightarrow\bf{[eqn.2]}$

• Putting the value of $\dfrac{Q_{1}}{Q_{2}}$ in [eqn.2]:-

$\dfrac{\sigma_{1}}{\sigma_{2}}=\dfrac{R_{2}}{R_{1}}$

• Put the value of R₁ and R₂

$\dfrac{\sigma_{1}}{\sigma_{2}}=\dfrac{10}{5}$

$\dfrac{\sigma_{1}}{\sigma_{2}}=\dfrac{2}{1}$

Therefore, The surface density of charge of the sphere will be in the ratio 2:1