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What is the electric field strength required to balance the gravitational force on a proton?

Physics

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Answer

1.02×10−7 N/C1.02 \times 10^{-7} \text{ N/C}

Step-by-step solution

4 steps

  1. Identify the given values and relevant physical constants.

    The mass of a proton is given as m=1.67×10−27 kgm = 1.67 \times 10^{-27} \text{ kg}.
    The charge of a proton is a fundamental constant, q=1.602×10−19 Cq = 1.602 \times 10^{-19} \text{ C}.
    The acceleration due to gravity is g=9.81 m/s2g = 9.81 \text{ m/s}^2.

  2. Set up the equation based on the problem statement.

    The problem asks for the electric field strength EE such that the electric force (FEF_E) on a proton is equal to its gravitational force (weight, FgF_g).
    The electric force is given by FE=qEF_E = qE.
    The gravitational force (weight) is given by Fg=mgF_g = mg.
    Equating these two forces:

    FE=FgF_E = F_g qE=mgqE = mg

  3. Solve the equation for the electric field strength EE.

    Divide both sides by qq:

    E=mgqE = \frac{mg}{q}

  4. Substitute the known values into the equation and calculate EE.

    E=(1.67×10−27 kg)×(9.81 m/s2)1.602×10−19 CE = \frac{(1.67 \times 10^{-27} \text{ kg}) \times (9.81 \text{ m/s}^2)}{1.602 \times 10^{-19} \text{ C}}

    First, calculate the numerator:

    1.67×10−27×9.81=16.3827×10−27 N1.67 \times 10^{-27} \times 9.81 = 16.3827 \times 10^{-27} \text{ N}

    Now, divide by the charge:

    E=16.3827×10−27 N1.602×10−19 CE = \frac{16.3827 \times 10^{-27} \text{ N}}{1.602 \times 10^{-19} \text{ C}} E=(16.38271.602)×10−27−(−19) N/CE = \left(\frac{16.3827}{1.602}\right) \times 10^{-27 - (-19)} \text{ N/C} E=10.2264×10−8 N/CE = 10.2264 \times 10^{-8} \text{ N/C} E=1.02264×10−7 N/CE = 1.02264 \times 10^{-7} \text{ N/C}

    Rounding to three significant figures:

    E≈1.02×10−7 N/CE \approx 1.02 \times 10^{-7} \text{ N/C}

    The electric field strength is:

    1.02×10−7 N/C1.02 \times 10^{-7} \text{ N/C}
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