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)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

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 so Φ=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}

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 = 0
so, Ф = 0 and flux , Ф = E.A = 0
so, E = 0
Hence, inside the sphere, the electric field equals zero.
(b) Take a Gaussian surface of radius r > R = 12cm
then, charged inclosed into the Gaussian surface is q = 1.6 × 10⁻⁷ C
So, Ф = q/ε₀
So, EA = q/ε₀
E = q/ε₀A, here A is the surface area of Gaussian spherical surface
e.g., A = 4πr²
So, E = q/4πε₀r² = 9 × 10⁹ × 1.6 × 10⁻⁷/(12 × 10⁻²)²
= 10⁵ N/C
(C) Similarly explanation of (B),
So, E = kq/r²
Here , k = 9 × 10⁹ Nm²/C² , q = 1.6 × 10⁻⁷C and r = 18cm
So, E = 9 × 10⁹ × 1.6 × 10⁻⁷/(18 × 10⁻²)²
= 4.44 × 10⁴ N/C.

(a) Radius of the spherical conductor, r = 12 cm = 0.12 m

Charge is uniformly distributed over the conductor, q = 1.6 × 10 ^-7 C

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 10⁵ N 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 10⁴ N/C