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Marrrta [24]
3 years ago
11

"Two identical positive charges exert a repulsive force of 6.7 × 10−9 N when separated by a distance 3.5 × 10−10 m. Calculate th

e charge of each. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of C."
Physics
1 answer:
Tamiku [17]3 years ago
8 0

Answer:

The charge of each charge is 3.02\times10^{-19} C

Explanation:

When yo have two charged particles they interact exerting an electrostatic force in the other particles, the magnitude of the electrostatic force between two particles is:

F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} (1)

with q1 and q2 the charges, r the distance between them and k the Coulomb's constant (k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})

Because the charges we're dealing are identical positive q1=q2, then (1) is:

F_{e}=k\frac{\mid q^2 \mid}{r^{2}}

Using the values the problem give us:

6.7\times10^{-9}=8.98755\times10^{9}\frac{\mid q^2 \mid}{(3.5\times10^{-10})^{2}}

solving for q:

q=\sqrt{\frac{(3.5\times10^{-10})^{2}6.7\times10^{-9}}{8.98755\times10^{9}}}

q= 3.02\times10^{-19} C

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