Question

Two spheres of equal charge densities have charges in the ratio 1 : 4 on their surfaces. Then the ratio of their volumes ????​

Answer 1

Given:

• The charge density of both the spheres is equal (σ)
• The charges of both the spheres are in the ratio 1: 4

Hence,

$\\ \blacksquare \: \sf{ \dfrac{q_1}{q_2} = \dfrac{1}{4}}$

Solution:

Charge density is given by,

$\dag \: \boxed{\sf{ \sigma = \dfrac{q}{A}}}$

here,

• σ = charge density
• q = charge
• A = area

As the charge density is the same for both the spheres,

$\longrightarrow \sf{ \dfrac{q}{A} = const }$

Hence,

$\longrightarrow \sf{ q \propto A }$

So,

$\longrightarrow \sf{\dfrac{q_1}{q_2} = \dfrac{A_1}{A_2} }$

_______________

We know that,

Area of a sphere (A)  = 4 πr²

Hence,

$\longrightarrow \sf{\dfrac{q_1}{q_2} = \dfrac{4 \pi {r_1}^2}{4\pi {r_2}^2} }$

$\longrightarrow \sf{\dfrac{q_1}{q_2} = \dfrac{ {r_1}^2}{{r_2}^2} }$

$\longrightarrow \sf{\dfrac{ {r_1}^2}{{r_2}^2} = \dfrac{1}{4} }$

$\longrightarrow \sf{\dfrac{ r_1}{r_2} = \sqrt{\dfrac{1}{4}}}$

$\longrightarrow \sf{\dfrac{ r_1}{r_2}= \dfrac{1}{2} }$

________________

The volume of the sphere is given by,

$\dag \: \boxed{\sf{ V = \dfrac{4}{3}\pi r^3}}$

The ratio of the volume of both the spheres will be,

$\longrightarrow \sf{ \dfrac{V_1}{V_2}= \dfrac{\dfrac{4}{3}\pi{ r_1}^3}{\dfrac{4}{3}\pi {r_2}^3}}$

$\longrightarrow \sf{ \dfrac{V_1}{V_2}= \dfrac{{ r_1}^3}{ {r_2}^3}}$

$\longrightarrow \sf{ \dfrac{V_1}{V_2}= {\bigg(\dfrac{1}{ 2}\bigg)}^3}$

$\longrightarrow \underline{\boxed{\sf{ \dfrac{V_1}{V_2}= \dfrac{1}{ 8}}}}$

The volume of the two spheres is in the ratio of 1:8.