Question

You have a 10 mm diameter teflon sphere. You charge it by rubbing it with a very small piece of cat fur so that it has a magnitude of 700 nC of charge on its surface. What uniform electric field magnitude (in N/C) would be needed to levitate the sphere against gravity

Answer 1

E = 5.04 10³ N / C

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Step-by-step explanation:

For the sphere to be in equilibrium the electric force must be equal to the weight of the sphere, we write the equilibrium equation

`F_(e)` - W = 0

q E = mg

E = mg / q

To find the mass of the Teflon sphere we use the density

ρ = m / V

m = ρ V

m = ρ 4/3 π r³

Teflon density is rho = 2.2 g / cm³ (1 kg / 10³ g) = 2.2 10⁻³ kg / cm³

The radius of the sphere

d = 10 mm = 1 cm

r = 0.5 cm

m = 2.2 10-3 4/3 π 0.5³

m = 3.6 10⁻⁴ kg

Let's calculate

E = 3.6 10⁻⁴ 9.8 / 700 10-9

E = 5.04 10³ N / C

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