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Calculate the electrostatic force between an alpha-particle and an electron separated by a distance of 10⁻¹³ m.

Physics

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Answer

4.61×10−2 N\text{4.61} \times 10^{-2} \text{ N}

Step-by-step solution

4 steps

  1. Identify the given values.

    The distance between the particles is r=10−13r = 10^{-13} m.
    The elementary charge is e=1.6×10−19e = 1.6 \times 10^{-19} C.
    Coulomb's constant is k=14πϵ0=9.0×109 Nm2/C2k = \frac{1}{4\pi\epsilon_0} = 9.0 \times 10^9 \text{ Nm}^2/\text{C}^2.

  2. Determine the charges of the alpha-particle and the electron.

    The charge of an electron is qe=−e=−1.6×10−19q_e = -e = -1.6 \times 10^{-19} C.
    An alpha-particle consists of two protons and two neutrons. The charge of an alpha-particle is qα=+2e=+2×(1.6×10−19 C)=3.2×10−19q_\alpha = +2e = +2 \times (1.6 \times 10^{-19} \text{ C}) = 3.2 \times 10^{-19} C.

  3. Apply Coulomb's Law to find the magnitude of the electrostatic force.

    Coulomb's Law states that the magnitude of the electrostatic force FF between two point charges q1q_1 and q2q_2 separated by a distance rr is given by:

    F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}

  4. Substitute the values and calculate the force.

    F=(9.0×109 Nm2/C2)∣(−1.6×10−19 C)×(3.2×10−19 C)∣(10−13 m)2F = (9.0 \times 10^9 \text{ Nm}^2/\text{C}^2) \frac{|(-1.6 \times 10^{-19} \text{ C}) \times (3.2 \times 10^{-19} \text{ C})|}{(10^{-13} \text{ m})^2} F=(9.0×109)(1.6×10−19)×(3.2×10−19)(10−13)2 NF = (9.0 \times 10^9) \frac{(1.6 \times 10^{-19}) \times (3.2 \times 10^{-19})}{(10^{-13})^2} \text{ N} F=(9.0×109)5.12×10−3810−26 NF = (9.0 \times 10^9) \frac{5.12 \times 10^{-38}}{10^{-26}} \text{ N} F=(9.0×109)×(5.12×10−38×1026) NF = (9.0 \times 10^9) \times (5.12 \times 10^{-38} \times 10^{26}) \text{ N} F=(9.0×109)×(5.12×10−12) NF = (9.0 \times 10^9) \times (5.12 \times 10^{-12}) \text{ N} F=46.08×109−12 NF = 46.08 \times 10^{9-12} \text{ N} F=46.08×10−3 NF = 46.08 \times 10^{-3} \text{ N} F=4.608×10−2 NF = 4.608 \times 10^{-2} \text{ N}

    The magnitude of the electrostatic force of attraction is:

    4.61×10−2 N\text{4.61} \times 10^{-2} \text{ N}
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