A parallel-plate capacitor is constructed of two disks spaced 2.00 mm apart. It is charged to a potential difference of 500. V. A proton

Question

A parallel-plate capacitor is constructed of two disks spaced 2.00 mm apart. It is charged to a potential difference of 500. V. A proton is shot through a small hole in the negative plate with a speed of 2.0 × 105 m/s. What is the farthest distance from the negative plate that the proton reaches?

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Nem 5 years 2021-08-27T20:04:53+00:00 1 Answers 32 views 0

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  1. thanhha
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    2021-08-27T20:06:16+00:00
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    Answer:

    0.000835\ \text{m}

    Explanation:

    d = Distance between plates = 2 mm

    V = Potential difference = 500 V

    v = Velocity of proton = 2\times 10^5\ \text{m/s}

    a = Acceleration

    m = Mass of proton = 1.67\times 10^{-27}\ \text{kg}

    Electric field is given by

    E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{500}{2\times 10^{-3}}\\\Rightarrow E=250000\ \text{V/m}

    Force balance is given by

    ma=qE\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 250000}{1.67\times 10^{-27}}\\\Rightarrow a=2.395\times 10^{13}\ \text{m/s}^2

    We have the relation

    v^2=u^2+2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(2\times 10^5)^2-0}{2\times 2.395\times 10^{13}}\\\Rightarrow s=0.000835\ \text{m}

    The farthest distance from the negative plate that the proton reaches is 0.000835\ \text{m}.

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