The electric field intensity 2m from the surface of a charged sphere is 30.86 x103 N/C. The charge density at the surface of the sphere is 8.85 x 19‒7 coul/m2. What is the radius of the sphere ?

Question

The electric field intensity 2m from the surface of a charged sphere is 30.86 x103 N/C. The charge density

at the surface of the sphere is 8.85 x 19‒7 coul/m2. What is the radius of the sphere ?

• Hint: The Gaussian surface area of the big sphere of radius, r + 2 is : A = 4π (r +2 )^2 Ans. r = 2.5 m

Answer 1

The key observation is that the electric field outside the sphere is the same

whether the charge is spread uniformly on the surface of the sphere, or

is concentrated at its center.

Let

r (to be calculated) be the radius of the sphere,

E = 30.86×10³ be the electric field 2m from the sphere,

d = 8.85×10^-7 C/m² be the density of the charge (I am assuming a typo in your statement),

k = 9×10⁹ kgm³/s²C² be the Coulomb constant,

Q = 4πr²d be the total change on the surface of the sphere.

If the change Q were at the center of the sphere, then the electric field 2m from the surface would be

kQ/(r+2)²

Thus we have the equation

E = kQ/(r+2)² = 4πkdr²/(r+2)²

You can solve this quadratic equation for r.

Notice, that if you plug in your numbers then

4πkd = 10⁵

and if you plug in r=2.5m, you will find that it is indeed a solution to the equation.