Question

Derive an expression for mechanical force per unit area of a charged conductor

Answer 1

Consider a small differential element dS on the surface of a charged conductor. Since the small differential element carries a charge and σ be the surface charge density;

dQ = σ* dS

Assume a point R which is located just outside the charge element. Then the electric intensity at that point R is;

E =σ/ε₀*k

Hence at R;

E = E₁+ E₂

E₁+ E₂ = σ/ε₀*k ⇒(1)

Since E₁ and E₂ are oppositely directed;  

E₁- E₂ =0

E₁ = E₂

Substituting the values in (1)

2*E₁ = σ/ε₀*k

E₁ = σ/2*ε₀*k

The repulsive forces experienced by the element dS carrying charge dQ;

F = [σ/2*ε₀*k]*σ* dS

F = σ²/2*ε₀ dS

The energy per unit volume or energy density;

dU/ dv = 1/2*ε₀*k*E²