A positive charge of 63.0 nC is 15.0 cm from a negative charge of 45.0 nC.The force on one of the charges due to the other is approximately A)1.13 $\times$ 10^-3 N B)1.13 $\times$ 10^-7 N C)1.02 $\times$ 10⁷ N D)1.25 $\times$ 10^-13 N E)1.02 $\times$ 10^-6 N

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The force between two charges is given by Coulomb's law:$$F = k\frac{q_1q_2}{r^2}$$where:* k is Coulomb's constant ($8.98755 × 10^9$ N m^2/C^2)* q_1 and q_2 are the charges of the two objects (in coulombs)* r is the distance between the two objects (in meters)In this problem, we have:* q_1 = 63.0 nC = $63.0 × 10^{-9}$ C* q_2 = -45.0 nC = $-45.0 × 10^{-9}$ C* r = 15.0 cm = 0.15 mPlugging these values into Coulomb's law, we get:$$F = (8.98755 × 10^9 	ext{ N m}^2/	ext{C}^2)\frac{(63.0 × 10^{-9} 	ext{ C})(-45.0 × 10^{-9} 	ext{ C})}{(0.15 	ext{ m})^2}$$$$F = 1.13 × 10^{-3} 	ext{ N}$$Therefore, the force on one of the charges due to the other is approximately 1.13 × 10^{-3} N.

A Positive Charge of 63 ×\times× 10-3 N B)1 ×\times× 10-7 N C)1

A Positive Charge of 63 $\times$ 10^-3 N
B)1 $\times$ 10^-7 N
C)1