Question

A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d = 35cm above charge Q1. The net force on Q2 is equal to zero.

a) Write an equation for the magnitude of charge Q2 in terms of the given variables.

b) Calculate the magnitude of Q2 in units of nanocoulombs

Answer 1

Answer:

Explanation:

Q1 = 35 nC = 35 x 10^-9 C

m = 3.5 micro gram = 3.5 x 10^-9 Kg

d  = 35 cm = 0.35 m

(a) The electrostatic force between the two charges is balanced by the weight of another charge.

F = m g

$\frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{d^{2}}=mg$

$Q_{2}=\frac{4\pi \epsilon _{0}mgd^{2}}{Q_{1}}$

(b) By substituting the values

$Q_{2}=\frac{3.5\times 10^{-9}\times 9.8\times 0.35\timees 0.35}{9\times 10^{9}\times 35\times 10^{9}}$

Q2 = 13.34 x 10^-12 C

Q2 = 0.0134 nC

Answer 2

The charge Q₂ is

(a) in terms of variables expressed as $Q_2=\frac{mgr^2}{kQ_1}$

(b) The magnitude of charge  Q₂ = 0.01334 nC

Given two charges Q₁ and Q₂ such that

the charge on Q₁ = 35 nC = 35×10⁻⁹ C

Mass of Q₂ is m = 3.5 microgram = 3.5×10⁻⁹ Kg

distance r between the charges = 35 cm = 0.35 m

(a) The charge Q₂ is floating because the electrostatic force F and the weight W of the charge are balanced, that is

F = W

$\frac{kQ_1Q_2}{r^2} =mg\\ Q_2=\frac{mgr^2}{kQ_1}$  

(b) The magnitude of Q₂:

$Q_2=\frac{3.5*10^{-9}*9.8*0.35*0.35}{9*10^9*35*10^{-9}}$

Q₂ = 13.34×10⁻¹² C

Q₂ = 0.01334 nC must be the magnitude of the charge so that it floats.

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