1) if a positive and negative charge each has a magnitude of 6 uC and are separated by a distance of 10 cm, what is the force between the particles?
2) Given that the repulsive force between two identical positive point charges is 1.0 N when the charges are 0.5 m apart, what is the value of each charge?
Show work please!
1) To calculate the force between two charges, you can use Coulomb's law formula:
F = k * ((q1 * q2) / r^2)
where F is the force between the charges, k is Coulomb's constant (9 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between them.
In this case, q1 = 6 uC (which is equivalent to 6 x 10^(-6) C), q2 = -6 uC, and r = 10 cm (which is equivalent to 0.1 m).
Replacing these values into the formula, we get:
F = (9 x 10^9 N m^2/C^2) * ((6 x 10^(-6) C) * (-6 x 10^(-6) C)) / (0.1 m)^2
Simplifying this equation, we have:
F = (9 x 10^9 N m^2/C^2) * (36 x 10^(-12) C^2) / 0.01 m^2
F = (9 x 10^9 N m^2/C^2) * (36 x 10^(-12) C^2) / 0.0001 m^2
F = 324 x 10^(-3) N
Therefore, the force between the particles is 324 mN (millinewtons).
2) To solve this problem, we can again use Coulomb's law formula:
F = k * ((q1 * q2) / r^2)
Here, we are given that F = 1.0 N, q1 = q2 = q (as both charges are identical), and r = 0.5 m.
Plugging in these values into the equation, we have:
1.0 N = (9 x 10^9 N m^2/C^2) * ((q * q) / (0.5 m)^2)
Simplifying this equation, we get:
1.0 N = (9 x 10^9 N m^2/C^2) * (q^2 / 0.25 m^2)
Rearranging the equation, we have:
q^2 = (1.0 N * 0.25 m^2) / (9 x 10^9 N m^2/C^2)
q^2 = 2.78 x 10^(-11) C^2
Taking the square root of both sides, we find:
q = √(2.78 x 10^(-11) C^2)
q ≈ 5.277 x 10^(-6) C
Therefore, the value of each charge is approximately 5.277 uC (microcoulombs).