Answer:
4.216 * 10^(-5) C
Explanation:
Parameters given:
Electrostatic force between charges, F = 4N
Distance between the charges, r = 2 m
The electrostatic force acting between two charged particles, q1 and q2, separated by a distance, r, is given as:
F = (k * q1 * q2) / r²
Where k = Coulombs constant.
Since q1 = q2 = q,
F = (k * q²) / r²
Inputting all the values:
4 = (9 * 10^9 * q²) / 2²
=> 16 = (9 * 10^9 * q²)
=> q² = 16 / (9 * 10^9)
q² = 1.78 * 10^(-9)
=> q = 4.216 * 10^(-5) C
The value of both charges is 4.216 * 10^(-5) C.
The values of each charge, when two identical charges, 2.0m apart, exert forces of magnitude 4.0N on each other is 23717 C
The question above can be solved using coulomb's law.
Coulomb's law: It states the force of attraction between two charges separated by a distance, is directly proportional to the product of the charges and inversely proportional to the square of the distance.
This can be expressed mathematically as
F = kq'q/r²................ Equation 1
Where F = force, q and q' are the two charges respectively, r = distance between the charges, k = coulombs constant.
Since: q = q', therefore
F = kq²/r².................. Equation 2
From the question,
Given: r = 2.0 m, F = 4.0 N, k = 9×10⁹ Nm/C²
Substitute these values into equation 2
4.0 = (9×10⁹)(q²)/2²
4 = (9×10⁹)(q²)/4
4×4 = (9×10⁹)(q²)
q² = 9×10⁹/16
q = √[(9/16)×10⁹]
q = 23717 C
Hence the value of either charge is 23717 C
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