Two identical balls of an electroscope, each having charges of 4 * 10- ^ 6 C, interact with a force of 1.6 N
| April 24, 2021 | education
Two identical balls of an electroscope, each having charges of 4 * 10- ^ 6 C, interact with a force of 1.6 N. At what distance are the centers of these balls?
Given: q (the magnitude of the charges of two identical balls) = 4 * 10-6 Cl = q1 = q2; F (force of interaction of two considered balls) = 1.6 N.
Constants: k (proportionality coefficient) = 9 * 10 ^ 9 N * m2 / Cl2.
The distance at which the centers of the balls will be located can be calculated using the Coulomb law: F = k * q1 * q ^ 2 / r ^ 2 = k * q ^ 2 / r ^ 2, whence r ^ 2 = k * q ^ 2 / F and r = √ (k * q ^ 2 / F).
Calculation: r = √ (9 * 10 ^ 9 * (4 * 10-6) ^ 2 / 1.6) = 0.3 m (30 cm).
Answer: The centers of the balls are located at a distance of 30 cm.
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