Respuesta :
Answer:
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