A charge of 1.3 * 10-9Kl in kerosene at a distance of 0.005m attracts a second charge with a force of 2 * 10-4N

A charge of 1.3 * 10-9Kl in kerosene at a distance of 0.005m attracts a second charge with a force of 2 * 10-4N. find the value of the second charge. The dielectric constant of kerosene is 2.

Data: q1 (first charge) = 1.3 * 10 ^ -9 C; r (distance) = 0.005 m; Fk (force of attraction of the second charge) = 2 * 10 ^ -4 N; ε (dielectric constant of kerosene) = 2; k (coefficient of proportionality) = 9 * 10 ^ 9 m / F.

We express the value of the second charge from the formula (we use Coulomb’s law): Fк = k * q1 * q2 / (ε * R ^ 2) and q2 = Fк * ε * R ^ 2 / (k * q1) = 2 * 10 ^ -4 * 2 * 0.005 ^ 2 / (9 * 10 ^ 9 * 1.3 * 10 ^ -9) = 8.55 * 10 ^ -10 Cl.

Answer: The second charge is 8.55 * 10 ^ -10 C.



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