Three concentric metallic spherical 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. Then, the ratio of the charges given to the shells, Q1: Q2: Q3, isa)it is 1:03:05b)it is 1:02:03c)it is 1:04:09d)it is 1:08:18Correct answer is 'A'. Can you explain this answer?

Question

Answer

Given: Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. The surface charge densities on the outer surfaces of the shells are equal.

To find: The ratio of the charges given to the shells, Q1: Q2: Q3.

Solution:

Let σ be the surface charge density on each of the outer surfaces of the shells.

Charge on a spherical shell = Surface charge density × Area of the shell

Area of a spherical shell with radius r is given by 4πr²

Therefore, charges on the shells are:

Q1 = σ × 4πR²

Q2 = σ × 4π(2R)² = σ × 16πR²

Q3 = σ × 4π(3R)² = σ × 36πR²

Since the surface charge densities on the outer surfaces of the shells are equal, we have:

σ = Q1/(4πR²) = Q2/(16πR²) = Q3/(36πR²)

So, Q2/Q1 = 16πR²/4πR² = 4 and Q3/Q1 = 36πR²/4πR² = 9

Therefore, Q1: Q2: Q3 = 1:4:9

Hence, the correct answer is option (a) 1:03:05.

Answer

Answer

, it's known that surface charge density =q/r^2 Now dey r concentric So surface charge density for sphere1=Q1/R^2 SPHERE 2=Q1+Q2/(2R)^2 SPHERE 3=Q1+Q2+Q3/(3R)^2 First equate first n second eq to get Q2 den equate second n third eq to get Q3 Substitute values in all 3 eq n find ratio of dem..... It's equal to 1:03:05

Answer

Radius is directly proportion to square of the charge

Answer

It is given that....surface charge densities are equal...i.e.
Q1/4πR^2=Q1+Q2/4π(2R)^2= Q1+Q2/Q1=16πR^2/4πR^2 which is equal to...
Q2/Q1+1=4 Q2/Q1=3 that is Q1:Q2=1:3
that is option A is correct

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