What must be the distance between point charge q_1=26.0 μC and point charge q_2=47.0 μC for the electrostatic force between them to have a magnitude of 5.70 N?

#### Solution:

Show me the final answer↓

We can use Coulomb’s law to figure out the distance between the two point charges. Coulomb’s law states:

F=k\dfrac{(q_1)(q_2)}{r^2}

(Where F is the electrostatic force, k is Coulomb’s constant, q_1 is the charge of the first particle, q_2 is the charge of the second particle, and r is the distance between the two charges)

Substitute the values we know into our equation.

F=k\dfrac{(q_1)(q_2)}{r^2}

5.7=(8.99\times 10^9)\dfrac{(26\times 10^{-6})(47\times 10^{-6})}{r^2}

5.7=(8.99\times 10^9)\dfrac{(26\times 10^{-6})(47\times 10^{-6})}{r^2}

(isolate for r)

5.7r^2=(8.99\times 10^9)(26\times 10^{-6})(47\times 10^{-6})

r^2=1.9273298

r=\sqrt{1.9273298}=1.39 m

#### Final Answer:

r=1.39 m

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