Find the electric field inside a hollow plastic ball of radius r that has charge q uniformly distributed on its outer surface. Give your answer as a multiple of q/ε0.

A) 0  for r < R B) Q/4πε₀r² for r > R

Explanation:

Here is the complete question

Part A Find the electric field inside a hollow plastic ball of radius R that has charge Q uniformly distributed on its outer surface. Give your answer as a multiple of Q/ε0.

Part B Find the electric field outside this ball. Give your answer as a multiple of Q/ε0. Express your answer in terms of some or all of the variables R, r and the constant π.

Solution

Using Gauss' law ∫E.dA = q/ε₀.  Where E is the electric field, dA is the area vector and q is the charge enclosed.

A For r < R The direction of the electric field is directed radially inward and r is outward and the angle between them is 180°. So E.dA = EdAcos180 = -EdA

∫-EdA = q/ε₀

-E∫dA =  q/ε₀

-E4πr² = q/ε₀ (∫dA = 4πr² since it is a sphere)

E = -q/4πr²ε₀

But for r < R q = 0. So,

E = -q/4πr²ε₀ = -0/4πr²ε₀ = 0

B For r > R The direction of the electric field is directed radially outward and r is outward and the angle between them is 0°. So E.dA = EdAcos0 = EdA

∫EdA = q/ε₀ where Q is the charge on the hollow plastic ball

E∫dA =  q/ε₀

E4πr² = q/ε₀  (∫dA = 4πr² since it is a sphere)

E = q/4πr²ε₀

But for r > R q = Q. So,

E = Q/4πr²ε₀ = Q/4πε₀r²

its c its is a mixture of gases