An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negati

Question

An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.

in progress 0
Nguyệt Ánh 5 months 2021-08-19T09:43:58+00:00 1 Answers 6 views 0

Answers ( )

    0
    2021-08-19T09:45:17+00:00

    Answer:

    Explanation:

    Given that  

    speed u=4×10⁶ m/s

    electric field E=4×10² N/c

    distance b/w the plates d=2 cm

    basing on the concept of the electrostatices

    now we find the acceleration b/w the plates

    acceleration a=qE/m=1.6×10⁻¹⁹×4×10²/9.1×10⁻³¹=0.7×10¹⁴=7×10¹³ m/s

    now we find the horizantal distance travelled by electrons hit the plates

    horizantal distance X=u[2y/a]^1/2

    =4×10⁶[2×2×10⁻²/7×10¹³]^1/2

    =9.5cm

    now we find the velocity f the electron strike the plate

    v²-(4×10⁶)²=2×7×10¹³×2×10⁻²

    v²=16×10¹²+28×10¹¹

    v²=1.88×10¹³m/s

    speed after hits =>V=4.34×10⁶ m/s

Leave an answer

Browse

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