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3 years ago

6

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

1 answer:

cupoosta [38]3 years ago

4 0

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

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