Two isolated conducting spheres of radii 2 cm and 8 cm are given initial charges of 5 C and 15 C respectively. They are then connected by a long, thin conducting wire. After electrostatic equilibrium is reached, what is the ratio of the electric field intensity just outside the surface of the smaller sphere to that of the larger sphere?

Step-by-step solution

When two conducting spheres are connected by a conducting wire, charge flows until their potentials become equal. V₁ = V₂ q₁ 4 ₀ r₁ = q₂ 4 ₀ r₂ The electric field just outside the surface of a conducting sphere is given by E = q 4 ₀ r^2 . This can be rewritten in terms of potential as E = V r . Since the spheres are at the same potential V , the ratio of their electric fields is: E₁ E₂ = V / r₁ V / r₂ = r₂ r₁ Given r₁ = 2 cm and r₂ = 8 cm , we get: E₁ E₂ = 8 2 = 4 1 The ratio of the electric field intensity of the