Two point charges, each with a charge of +1.00 Coulomb are separated by a distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between them.a.- 9.0 x 107 J.b.9.0 x 109 J.c.- 9.0 x 109 J.d.9.0 x 107 J.
Question
Two point charges, each with a charge of +1.00 Coulomb are separated by a distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between them.a.- 9.0 x 107 J.b.9.0 x 109 J.c.- 9.0 x 109 J.d.9.0 x 107 J.
Solution
The force between two charges is given by Coulomb's Law:
F = k * |q1*q2| / r^2
where:
- F is the force between the charges,
- k is Coulomb's constant (8.99 x 10^9 N.m^2/C^2),
- q1 and q2 are the charges (in this case, both are +1.00 C), and
- r is the distance between the charges (1.00 m in this case).
Substituting the given values into the equation, we get:
F = (8.99 x 10^9 N.m^2/C^2) * (1.00 C * 1.00 C) / (1.00 m)^2 F = 8.99 x 10^9 N
So, the magnitude of the electrical force of repulsion between the two charges is 8.99 x 10^9 N. However, none of the options exactly match this result. The closest option is b. 9.0 x 10^9 J, but note that the units are incorrect. The force should be in Newtons (N), not Joules (J).
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