What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?

Physics

1 answer:

Nastasia [14]3 years ago

8 0

Answer:

$F=1.26*10^{-3}N$

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

$F=\frac{kq_1q_2}{d^2}$

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges( $q_1,q_2$ ) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N$

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