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3 years ago

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Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19

C and Coulomb’s constant is 8.987 × 109 N · m2/C2. Answer in units of m.

1 answer:

sergejj [24]3 years ago

5 0

The distance between the charges is 13.86 X 10⁴m

<u>Explanation:</u>

Given:

Force, F = 1.2N

Charge, q₁ = 1.602 X 10⁻¹⁹ C

k = 8.987 X 10⁹ Nm²/C²

Distance, d = ?

According to Coulomb's law:

$F = k\frac{q_1q_2}{r^2} \\\\$

Substituting the value in the formula we get:

$1.2 = 8.987X 10^9 X\frac{1.602 X 10^-^1^9 X 1.602 X 10^1^9}{r^2} \\\\1.2 = \frac{23.06 X 10^9}{r^2} \\\\r^2 = 19.22 X 10^9\\\\r = 13.86 X 10^4m$

Therefore, the distance between the charges is 13.86 X 10⁴m

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