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3 years ago

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"Two electrostatic point charges of +68.0 µC and +56.0 µC exert a repulsive force on each other of 179 N. What is the distance b

etween the two charges? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of m."

2 answers:

timama [110]3 years ago

5 0

Answer: The distance between the charges is r=0.44m

Explanation

Step one

Using the formula

F = (k q1*q2) /r²

F = electric force

k = Coulomb constant

q1, q2 = charges

r = distance of separation

Step two

Given data

F=179N

q1 and q2=+68.0 µC and +56.0 µC

k=8.98755 × 10^9 N

r=?

Step three

Inputting our values we have

179=8.9875×10^9(68*10^-6*56*10^-6)/r²

179=(8.9875*10^9*3.8*10^-9)r²

179=34.15/r²

Make r subject of formula we have

r²=34.15/179

r²=0.190

r= √0.190

r=0.44m

Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Svetradugi [14.3K]3 years ago

5 0

Answer:

r = 0.437m≈0.4m

Explanation:

F= \frac{KQq}{r²}

Where K= 8.98755 × 109 N · m2 /C 2

F= 179 N

Q= +68.0 µC= 68.0 × 10^{-6}C

q= +56.0 µC= 56.0 × 10^{-6}C

r= ?

To find r, make r the subject of the formula.

F= \frac{KQq}{r²}

r²= \frac{KQq}{F}

r= \sqrt{} \frac{KQq}{F}

Then input the values,

r = \sqrt{} \frac{(8.98755 X 10^{9} )(68.0 X 10^{-6} )(56.0 X 10^{-6} ) }{179}

= \sqrt{} \frac{34224.5904 X 10^{9-12} }{179}

= \sqrt{191.1988 X 10^{-3}

= 0.437m≈0.4m

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