Two 10¢ coins (dimes) carrying identical charges are lying 2.5 m apart on a table....

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?  Blank 1. Fill in the blank, read surrounding text. C.  .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?  Blank 1. Fill in the blank, read surrounding text. ______C.  

Answer 1

It seems the question here was duplicated. The charge (in Coulombs, C) on each coin is unknown, but the same, so we will call each Q. The magnitude of the electrostatic force experienced by each coin is given by Coulomb's Law:

F = k*Q₁*Q₂/r²

Where k is the Coulomb constant (approximately 9.0*10⁹ N*m²/C²) and r is the distance between them in meters. We are given this distance (2.5 m) and the apparent force (2.0 N) and we know Q₁=Q₂=Q, so we can substitute these known quantities and solve for Q:

(2.0) = (9.0*10⁹)Q²/(2.5)² → Q = √[(2.0)*(2.5²)/(9.0*10⁹)] = 3.7*10^-5 C

Without additional context, we do not know whether these charges are both positive or both negative (which also arises mathematically from the square root), but the quantity has been obtained.