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Two coins lie 1.5 meters apart on a table. They carry identical electric charges. Approximately how
large is the charge on each coin if each coin experiences a force of 2.0 N

Sagot :

I don’t have any of the problem I can get it to my computer for my work
Lanuel

The charge on each coin is equal to [tex]2.50 \times 10^{-10}\;Coulomb[/tex]

Given the following data:

  • Force = 2.0 Newton
  • Radius = 1.5 meters

To determine the charge on each coin, we would apply the law of electrostatic forces:

Mathematically, the law of electrostatic forces is given by the formula:

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

Note: [tex]q_1 = q_2[/tex] (since the coin carry identical electric charges).

Therefore, the formula becomes:

[tex]F = \frac{k(2q)}{r^2}[/tex]

Making q the subject of formula, we have:

[tex]q=\frac{Fr^2}{2k}[/tex]

Substituting the given parameters into the formula, we have;

[tex]q=\frac{2 \times 1.5^2}{2 \times 8.99 \times 10^9}\\\\q=\frac{2 \times 2.25}{2 \times 8.99 \times 10^9}\\\\q=\frac{4.5}{17.98 \times 10^9}\\\\q=2.50 \times 10^{-10}\;Coulomb[/tex]

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