A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative plate with a s

Question

A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative plate with a speed of 3.0 x 105m/s. What is its speed when it emerges through the hole in the positive plate? (Hint: The electric potential outside of a parallel-plate capacitor is zero).Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative plate with a m/s. What is its speed when it emerges through the hole in the positive plate? (Hint: The electric potential outside plate capacitor is

in progress 0
Ngọc Khuê 13 hours 2021-07-22T00:59:11+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-22T01:01:08+00:00

    Answer:

    The speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

    Explanation:

    Given that,

    A parallel-plate capacitor is held at a potential difference of 250 V.

    A A proton is fired toward a small hole in the negative plate with a speed of, u=3\times 10^5\ m/s

    We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

    qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s

    So, the speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )