A uniformly charged conducting sphere of diameter 3.5 cm has a surface charge density of 20 C m ⁻² . The total electric flux leaving the surface of the sphere is nearly [permittivity of free space, ₀=8.85 10⁻¹² SI unit]

Step-by-step solution

Total electric flux through any closed surface equals the enclosed charge divided by the permittivity of free space: _E = Q / ₀ . The sphere has surface charge density = 20 10⁻⁶ , C ,m ⁻² and radius R = 1.75 10⁻² , m . The enclosed charge is Q = A = 4 R^2 . Substituting numerical values: Q = (20 10⁻⁶) 4 (1.75 10⁻²)^2 = 245 10⁻¹⁰ , C The electric flux becomes: _E = 245 10⁻¹⁰ 8.85 10⁻¹² = 245 8.85 10^2 , N m^2/C Evaluating numerically with 3.14159 : _E 769.69 8.85 10^2 8.7 10^3 , N m^2/C This value corresponds to opt