Prove electric field=(surface charge density ×radius)÷(3×permitivity of free space)?

To prove this equation, we can start with the formula for the electric field due to a uniformly charged sphere:

E = (1/(4πε₀)) * (Q / r²)

Where:
E = electric field
ε₀ = permittivity of free space
Q = total charge of the sphere
r = radius of the sphere

We can rewrite the total charge of the sphere in terms of surface charge density (σ) and radius (r):

Q = 4πr²σ

Substitute Q into the equation for the electric field:

E = (1/(4πε₀)) * ((4πr²σ) / r²)

Simplify the equation:

E = (1/(4πε₀)) * (4σ)

E = σ / ε₀

Therefore, the electric field due to a uniformly charged sphere is equal to the surface charge density divided by the permittivity of free space.

Consider a uniformly charged spherical surface of radius 'R' and surface charge density 'σ'.

The electric field (E) at a point 'P' outside the sphere (r > R) is given by:

E = k - Q/r²

where k is Coulomb's constant, Q is the total charge on the sphere, and r is the distance from the center of the sphere to point 'P'.

The total charge on the sphere is given by:

Q = σ - 4πR²

Substituting this expression for Q into the equation for E, we get:

E = k - σ - 4πR² / r²

Now, the surface charge density (σ) is related to the electric field (E) by the following equation:

σ = (E - ε₀) / 2

where ε₀ is the permittivity of free space.

Substituting this expression for σ into the equation for E, we get:

E = k - (E - ε₀) - 4πR² / (2 - r²)

Simplifying this equation, we get:

E = (k - 2ε₀ - 4πR²) / (3 - r²)

Substituting k = 1/(4πε₀) into this equation, we get:

E = (1/(4πε₀)) - (2ε₀ - 4πR²) / (3 - r²)

Simplifying further, we get:

E = (R - σ) / (3ε₀)

Therefore, the electric field (E) outside a uniformly charged spherical surface is given by the equation:

E = (σ - R) / (3ε₀)