Answer:
a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
Explanation:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by
E = s/2ε₀[1 - z/√(z² + R²)]
So, the net force on the small plastic sphere of mass M and charge Q is
F = QE
F = Qs/2ε₀[1 - z/√(z² + R²)]
b. At what height h does the sphere hover?
The sphere hovers at height z = h when the electric force equals the weight of the sphere.
So, F = mg
Qs/2ε₀[1 - z/√(z² + R²)] = mg
when z = h, we have
Qs/2ε₀[1 - h/√(h² + R²)] = mg
[1 - h/√(h² + R²)] = 2mgε₀/Qs
h/√(h² + R²) = 1 - 2mgε₀/Qs
squaring both sides, we have
[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²
h²/(h² + R²) = (1 - 2mgε₀/Qs)²
cross-multiplying, we have
h² = (1 - 2mgε₀/Qs)²(h² + R²)
expanding the bracket, we have
h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²
collecting like terms, we have
h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²
Factorizing, we have
[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²
So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]
taking square-root of both sides, we have
√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]
h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
