A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field (a) Inside the sphere (b) Just outside the sphere (c) At a point 18 cm from the centre of the sphere?

a) We know, charge have nature to reside outer surface of the conductor. It means, charge inside the surface equals zero.According to Gaussian theorem,Ф = q/ε ₀, here q is charged inclosed the Gaussian surface.∵ q = 0so, Ф = 0 and flux , Ф = E.A = 0so, E = 0Hence, inside the sphere, the electric field equals zero.(b) Take a Gaussian surface of radius r > R = 12cmthen, charged inclosed into the Gaussian surface is q = 1.6 × 10⁻⁷ CSo, Ф = q/ε₀So, EA = q/ε₀E = q/ε₀A, here A is the surface area of Gaussian spherical surfacee.g., A = 4πr²So, E = q/4πε₀r² = 9 × 10⁹ × 1.6 × 10⁻⁷/(12 × 10⁻²)²= 10⁵ N/C(C) Similarly explanation of (B),So, E = kq/r²Here , k = 9 × 10⁹ Nm²/C² , q = 1.6 × 10⁻⁷C and r = 18cmSo, E = 9 × 10⁹ × 1.6 × 10⁻⁷/(18 × 10⁻²)²= 4.44 × 10⁴ N/C. read less

(a)Radius of the spherical conductor, r = 12 cm = 0.12 m Charge is uniformly distributed over the conductor, q = 1.6 × 10-7C Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it. (b)Electric field E just outside the conductor is given by the relation, Where, = Permittivity of free space Therefore, the electric field just outside the sphere is 105N C-1 (c)Electric field at a point 18 m from the centre of the sphere = E1 Distance of the point from the centre, d = 18 cm = 0.18 m Therefore, the electric field at a point 18 cm from the centre of the sphere is . 4.4 x 104N/C read less

(a)electric field inside a spherical conductor is zero. It is obvious that if we consider a Gaussian surface inside sphere it will enclose zero charge because all charges reside on the surface. (b) Electric field just outside the surface will be E= q/4π∈0r² where r= 12 cm . Putting these values we get E=105N (c) Similarly for r=18 cm we get E=44.44×105N read less

According to gauss law, Φ=q/e Φ=E.A( E=electric field and A=area enclosed by the Gaussian surface) where q is net charge enclosed by the Gaussian surface and e is the permitivity of free space(8.85*10-12) Now, a) inside the sphere charge is zero soΦ=0 and E is also zero ( from above b) apply above formula with area of sphere =4Π(r)2 and find electric field C) same as b case but here the radius changes to 18cm .Apply gauss law read less