Question

A conducting sphere of radius 10 cm is charged with 10 microcoulomb another uncharged fare of radius 20cm is allowed to touch it for some time after that if the spheres are separated then surface density of charge of the sphere will be in the ratio of

Answer 1

When the two conducting spheres touch each other there will be a flow of charge until they both have the same potential.

Let R1 and R2 be the radii of spheres 1 and 2, respectively.  Let Q1 and Q2 be the charges on spheres 1 and 2, respectively, after they are separated. Since both have the same potential,

V = Q1/C1 = Q2/C2

  ⇒ Q1/(4πϵ0R1) = Q2/(4πϵ0R2)

  ⇒ Q1/R1 = Q2/R2

  ⇒  Q1/Q2 = R1/R2

Surface charge density on sphere 1 = σ1 =  Q1/(4πR12)

Surface charge density on sphere 2 = σ2 = Q2/(4πR22)

σ1 /σ2 = (Q1/Q2) x (R22/R12)  

          =  (R1/R2) x (R22/R12)  

          = R2/R1

          = 20 cm/10 cm

          = 2

Therefore   σ1 : σ2 = 2 : 1

Answer 2

Answer:

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

Explanation:

Given that,

Radius of sphere = 10 cm

Charge =10 micro coulomb

Radius of uncharged sphere = 20 cm

When the two conducting spheres touch each other there will be a flow of charge until they both have the same potential .

Let R₁ and R₂ be the radius of  both sphere respectively

Let Q₁ and Q₂  be the charge on both sphere respectively.

Let the common potential = V

The charge on first sphere,

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

The charge on second sphere,

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

The ratio of the charge of spheres,

$\dfrac{Q_{1}}{Q_{2}}=\dfrac{R_{1}}{R_{2}}$.....(I)

Now, The charge density on the first sphere

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

The charge density on the second sphere

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

The ratio of the charge density

$\dfrac{\sigma_{1}}{\sigma_{2}}=\dfrac{Q_{1}R_{2}^2}{Q_{2}R_{1}^2}$..(II)

Put the value of $\dfrac{Q_{1}}{Q_{2}}$ in equation (II)

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

Put the value of R₁ and R₂

$\dfrac{\sigma_{1}}{\sigma_{2}}=\dfrac{20}{10}$

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

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