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3 years ago

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Part A By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on e

ach object is 4.0 N ? Express your answer with the appropriate units.
Part B How much charge should be placed on each of two small objects separated by 1.0 m so that the electric force exerted on each is 4.0 N ? Express your answer with the appropriate units.

1 answer:

aalyn [17]3 years ago

6 0

Answer:

Part A: <em>4.74×10⁴ m</em>

P<em>art </em>B<em>: </em>0.000211 C

Explanation:

Part A:

From coulombs law,

F = kq²/r²........................ Equation 1

Note: The two object are carrying equal charges.

Where F = Electric force, q = charge, r = distance of separation, k = proportionality constant

<em>Given: F = 4.0 N, q = 1.0 C, </em>

<em>Constant: k = 1/4πe₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 1,</em>

<em>4 = 9×10⁹(1²)/r²</em>

<em>r² = (9×10⁹)/4</em>

<em>r² = 2.25×10⁹</em>

<em>r = √(2.25×10⁹)</em>

<em>r = 4.74×10⁴ m.</em>

<em>Part 2:</em>

<em>F = kq²/r²</em>

<em>Making q the subject of the equation</em>

<em>q = √(Fr²/k)........................... Equation 2</em>

Where: F = 4.0 N, r = 1.0 m, k = 9×10⁹ Nm²/c²

Substituting these values into equation 2

q = √(4×1²/(9×10⁹ )

q = 2.11/10⁴

q = 0.000211 C.

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