Question

3.3.2 test (cst): forces
question 8 of 20
one particle has a charge of 4.2×10⁻⁹ c, while another particle has a charge of 1.10×10⁻⁹ c. if the two particles are separated by 0.005 m, what is the electromagnetic force between them? the equation for coulombs law is ( f_e=\frac{kq_1q_2}{r^2} ), and the constant, k, equals 9.00×10⁹ n·m²/c².
a. 1.85×10⁻¹³ n
b. 1.66×10⁻³ n
c. 1.66×10⁻⁵ n
d. 8.31×10⁻⁶ n

Explanation:

Step1: Substitute values into Coulomb's law formula

Given \(k = 9.00\times10^{9}\space N\cdot m^{2}/C^{2}\), \(q_{1}=4.2\times 10^{-9}\space C\), \(q_{2}=1.10\times 10^{-9}\space C\), \(r = 0.005\space m\).

$$ F_{e}=\frac{kq_{1}q_{2}}{r^{2}}=\frac{(9.00\times 10^{9})(4.2\times 10^{-9})(1.10\times 10^{-9})}{(0.005)^{2}} $$

Step2: Calculate numerator and denominator

• Numerator: \((9.00\times 10^{9})(4.2\times 10^{-9})(1.10\times 10^{-9})=9\times4.2\times1.1\times10^{9 - 9-9}=41.58\times10^{-9}\)
• Denominator: \((0.005)^{2}=2.5\times10^{-5}\)

Step3: Calculate the force

$$ F_{e}=\frac{41.58\times 10^{-9}}{2.5\times 10^{-5}}=\frac{41.58}{2.5}\times10^{-9 + 5}=16.632\times10^{-4}=1.6632\times 10^{-3}\space N $$

Answer:

B. \(1.66\times 10^{-3}\space N\)