Two electric charges, one of which is half the size of the other, being in a vacuum at a distance of 0.6 m

| February 11, 2021 | education

Two electric charges, one of which is half the size of the other, being in a vacuum at a distance of 0.6 m, interact with a force of 2 mN. Determine these charges.

q1 = q2 / 2.
r = 0.6 m.
F = 2 mN = 2 * 10 ^ -3 N.
k = 8.98 * 10 ^ 9 N * m ^ 2 / Kl ^ 2.
q -?
According to Coulomb’s law, the force of interaction between two charges is determined by the formula: F = k * q1 * q2 / r ^ 2.
Where k is the proportionality coefficient, q1, q2 are the values of the interacting charges, r is the distance between the charges.
In our case, for q1 = q2 / 2, the formula will look like: F = k * q1 * 2 * q1 / r ^ 2 = 2 * k * q1 ^ 2 / r ^ 2.
Let us express the amount of charge: q1 ^ 2 = F * r ^ 2 / k.
q1 = √ (F * r ^ 2/2 * k).
q1 = √ (2 * 10 ^ -3 H * (0.6 m) ^ 2/2 * 8.98 * 10 ^ 9 H * m ^ 2 / Cl ^ 2) = √0.04 * 10 ^ -12 = 0.2 * 10 ^ -6 Cl.
q2 = 2 * q1.
q2 = 2 * 0.2 * 10 ^ -6 Cl = 0.4 * 10 ^ -6 Cl.
Answer: q1 = 0.2 * 10 ^ -6 C, q2 = 0.4 * 10 ^ -6 C.



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