Question

Three isolated metal spheres A, B, C have radius R, 2R, 3R respectively, and same charge Q. U_A, U_B and U_C be the energy density just outside the surface of the spheres. The relation between U_A, U_B and U_C is

Answer 1

The energy density just outside the surface of a charged sphere is given by the formula:
U_A = \(\frac {1}{2}\)ε₀E².
The electric field outside a uniformly charged sphere is given by:
E = \(\frac {kQ}{r^2}\)
Let's consider the three spheres A, B, and C with radii R, 2R, and 3R, respectively.
For sphere A (radius R):,ṁ
E_A = \(\frac {kQ}{R^2}\)
U_A = \(\frac {1}{2}\)ε₀\((\frac {kQ}{R^2})^2\) = \(\frac {1}{2}\)ε₀\(\frac {k^2Q^2}{R^4}\)
For sphere B (radius 2R):
E_B = \(\frac {kQ}{(2R)^2}\) = \(\frac {kQ}{4R^2}\)
U_B = \(\frac {1}{2}\)ε₀\((\frac {kQ}{4R^2})^2\) = \(\frac {1}{2}\)ε₀\(\frac {k^2Q^2}{16R^4}\) =\(\frac {1}{8}\)ε₀\(\frac {k^2Q^2}{R^4}\)
For sphere C (radius 3R):
E_C = \(\frac {kQ}{(3R)^2}\) = \(\frac {kQ}{9R^2}\)
U_C = \(\frac {1}{2}\)ε₀\((\frac {kQ}{9R^2})^2\) = \(\frac {1}{2}\)ε₀\(\frac {kQ}{81R^4}\) = \(\frac {1}{18}\)ε₀\(\frac {k^2Q^2}{R^4}\)
Therefore, the relation between U_A, U_B, and U_C is: 
U_A : U_B : U_C = 1 : \(\frac {1}{8}\) : \(\frac {1}{18}\)
U_A : U_B : U_C = 18 : 2 : 1.
Hence, the relation between U_A, U_B, and U_C is 18 : 2 : 1 i.e. U_A>U_B>U_C

Answer 2

The electric energy density (u) is given by:
u = (1/2) ε E²
For a charged sphere, the electric field just outside the surface is:
E = q / (4πεr²)
For sphere A (radius r):
E = q / (4πεr²)
u_a = q² / (32π²εr⁴)
For sphere B (radius 2r):
E = q / (16πεr²)
u_b = q² / (512π²εr⁴)
For sphere C (radius 3r):
E = q / (36πεr²)
u_c = q² / (2592π²εr⁴)
Ratio of energy densities:
u_b / u_a = 16
u_c / u_a = 81
Relation between energy densities:
u_b : u_a : u_c = 16 : 1 : 81