Sorry....my physics teacher used the constant value of k to be 9×10⁹

Physics

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Answer

2.31×10−8 N\mathbf{2.31 \times 10^{-8} \text{ N}}

Step-by-step solution

3 steps

Here is the calculation for part a) using k=9×109 N m2/C2k = 9 \times 10^9 \text{ N m}^2/\text{C}^2.

a) Calculate the magnitude of the electrostatic force between an electron and a proton.
Given:

  • Coulomb's constant, k=9×109 N m2/C2k = 9 \times 10^9 \text{ N m}^2/\text{C}^2
  • Elementary charge (magnitude), e=1.602×10−19e = 1.602 \times 10^{-19} C
  • Charge of electron, qe=−1.602×10−19q_e = -1.602 \times 10^{-19} C
  • Charge of proton, qp=+1.602×10−19q_p = +1.602 \times 10^{-19} C
  • Distance, r=1.0×10−10r = 1.0 \times 10^{-10} m (from the original problem statement)

  1. Use Coulomb's Law to find the magnitude of the electrostatic force.

    F=k∣qeqp∣r2F = k \frac{|q_e q_p|}{r^2}

  2. Substitute the given values into the formula.

    F=(9×109 N m2/C2)∣(−1.602×10−19 C)(+1.602×10−19 C)∣(1.0×10−10 m)2F = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-1.602 \times 10^{-19} \text{ C})(+1.602 \times 10^{-19} \text{ C})|}{(1.0 \times 10^{-10} \text{ m})^2}

  3. Calculate the force.

    F=(9×109)(1.602×10−19)2(1.0×10−10)2 NF = (9 \times 10^9) \frac{(1.602 \times 10^{-19})^2}{(1.0 \times 10^{-10})^2} \text{ N} F=(9×109)2.566404×10−381.0×10−20 NF = (9 \times 10^9) \frac{2.566404 \times 10^{-38}}{1.0 \times 10^{-20}} \text{ N} F=(9×109)×(2.566404×10−18) NF = (9 \times 10^9) \times (2.566404 \times 10^{-18}) \text{ N} F=23.097636×10−9 NF = 23.097636 \times 10^{-9} \text{ N} F=2.3097636×10−8 NF = 2.3097636 \times 10^{-8} \text{ N}

    Rounding to three significant figures:

    F≈2.31×10−8 NF \approx \boxed{\mathbf{2.31 \times 10^{-8} \text{ N}}}
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