Question

Two 10¢ coins (dimes) carrying identical charges are lying 2.5 m apart on a table. If each of these coins experiences an electrostatic force of magnitude 2.0 N due to the other coin, how large is the charge on each coin?

I know the answer is 52 micro coulombs, but I keep getting 37.26...

Answer 1

Charge q = 37.27 micro coulombs

Step-by-step explanation:

Given:

q₁ = q₂ = q

r = 2.5 m

F = 2.0 N

________

q - ?

Coulomb's law:

F =k·q₁·q₂ / r² = k·q·q / r² = k·q² / r²

Charge:

q = √ (F·r² / k)

q = √ (2·2.5² / (9·10⁹) ≈ 37.27·10⁻⁹ C

or q = 37.27 micro coulombs

The correct answer is not 52, but !!! :))