Answer to Question #195788 in Physics for Salvador

Question

Two charges each with a magnitude of1.1

â‹…

1

−

7

1.1⋅10−7

C experience a force of 4.2

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1

−

4

4.2⋅10−4

N. How far apart are the charges from each other? 

Answer 1

According to the Coulomb's law, the force experienced by each charge is:

$$F = k\dfrac{q^2}{r^2}$$

where $q = 1.1\times 10^{-7}C$ is the charge of both charges, $r$ is the distance between charges, and $k = 9\times10^{9}\space N\cdot m^2/C^2$ is the Coulomb's constant. Expressing $r$ and substituting $F = 4.2\times 10^{-4}N$, obtain:

$$r = q\sqrt{\dfrac{k}{F}}\\ r = 1.1\times 10^{-7}\cdot \sqrt{\dfrac{9\times10^{9}}{4.2\times 10^{-4}}} \approx 0.51m$$

Answer. 0.51 m.