Two identical point charges, located in the oil at a distance of 6 cm from each other, interact with

| January 31, 2021 | education

Two identical point charges, located in the oil at a distance of 6 cm from each other, interact with a force of 4 * 10 ^ -4 N. What are these charges if the dielectric constant of the oil is 2.5?

To find out the value of each of the indicated point charges, we use the formula: Fк = k * q ^ 2 / (εм * r ^ 2), whence we express: q = √ (Fк * εм * r ^ 2 / k).

Const: εm is the dielectric constant of the oil (according to the condition εm = 2.5); k – coefficient of proportionality (k = 9 * 10 ^ 9 N * m2 / Kl2).

Data: Fк – force of interaction of charges (Fк = 4 * 10 ^ -4 N); r – distance (r = 6 cm; in SI system r = 6 * 10 ^ -2 m).

Let’s perform the calculation: q = √ (Fk * εm * r ^ 2 / k) = √ (4 * 10 ^ -4 * 2.5 * (6 * 10 ^ -2) ^ 2 / (9 * 10 ^ 9)) = 2 * 10 ^ -8 Cl.

Answer: Each of the indicated point charges is 20 nC.



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