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3 years ago

The force between a pair of .005 charges is 750 N. What is the distance between them?

Physics

1 answer:

Ganezh [65]3 years ago

6 0

Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?

Answer:

17.32 m

Explanation:

From coulomb's Law,

F = kqq'/r²........................... Equation 1

Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.

make r the subject of the equation above

r = √(kqq'/F)..................... Equation 2

From the question,

Given: q = q' = 0.005 C, F = 750 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(9.0×10⁹×0.005×0.005/750)

r = √(300)

r = 17.32 m.

Hence the distance between the pair of charges = 17.32 m

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Answer:

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D₁: Displacement Vector 1

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D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

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