Three concentric metallic sphere shells of radii R,2R,3R are given charges Q1,Q2,Q3 respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. What is the ratio of the charges given to the shells Q1,Q2,Q3?

sigma=Q/area

but areas are proportioal to radius squared.

so if the sigmas are equal, then

Q1/r1^2=Q2/r2^2=Q2/r3^2

I assume you can take it from there.

1:4:9

To solve this problem, we can use the concept of charge distribution on conductive surfaces.

Given:
Radius of the inner sphere shell (r1) = R
Radius of the middle sphere shell (r2) = 2R
Radius of the outer sphere shell (r3) = 3R

Charge on the inner shell (Q1) = ?
Charge on the middle shell (Q2) = ?
Charge on the outer shell (Q3) = ?

Surface charge density on the outer surfaces of all the shells is equal.

Surface charge density on the outer surface of the inner shell (𝜎1) = Surface charge density on the outer surface of the middle shell (𝜎2) = Surface charge density on the outer surface of the outer shell (𝜎3)

To find the ratio of the charges given to the shells, we need to relate the charges to the surface charge densities.

The surface charge density on a conducting sphere can be given by 𝜎 = Q / (4𝜋r^2), where Q is the charge and r is the radius of the sphere.

So, we can equate the surface charge densities on the outer surfaces of the shells:

𝜎1 = 𝜎2 = 𝜎3

Q1 / (4𝜋r1^2) = Q2 / (4𝜋r2^2) = Q3 / (4𝜋r3^2)

Q1 / (r1^2) = Q2 / (r2^2) = Q3 / (r3^2)

Substituting the values of r1, r2, and r3:

Q1 / R^2 = Q2 / (2R)^2 = Q3 / (3R)^2

Simplifying:

Q1 / R^2 = Q2 / 4R^2 = Q3 / 9R^2

Comparing the first and second ratios, we can write:

Q1 : Q2 = 1 : 4

Comparing the second and third ratios, we can write:

Q2 : Q3 = 4 : 9

Therefore, the ratio of the charges given to the shells Q1, Q2, Q3 is 1 : 4 : 9.

To find the ratio of the charges given to the shells Q1, Q2, and Q3, we need to start by first understanding the concept of surface charge density.

Surface charge density (σ) is defined as the total charge (Q) distributed over the surface area (A) of an object. Mathematically, it is represented as σ = Q/A.

In this problem, the surface charge densities on the outer surfaces of the shells are equal. Let's denote this common surface charge density as σ.

The surface area of a sphere is given by A = 4πr^2, where r is the radius of the sphere. So, we can express the surface charge densities of the given shells as:

Shell 1 (with radius R): σ1 = Q1 / (4πR^2)
Shell 2 (with radius 2R): σ2 = Q2 / (4π(2R)^2) = Q2 / (16πR^2)
Shell 3 (with radius 3R): σ3 = Q3 / (4π(3R)^2) = Q3 / (36πR^2)

Since it is given that the surface charge densities on the outer surfaces of the shells are equal (σ1 = σ2 = σ3 = σ), we can set up the following equations:

σ1 = σ
σ2 = σ
σ3 = σ

Substituting the expressions for σ1, σ2, and σ3, we have:

Q1 / (4πR^2) = Q2 / (16πR^2) = Q3 / (36πR^2)

Canceling out the common factor, we get:

Q1 = Q2 / 4
Q2 = Q3 / 9

Therefore, the ratio of the charges given to the shells Q1, Q2, Q3 is:

Q1 : Q2 : Q3
1 : 4 : 9