CALLING ALL PHYSICS NERDS!!!! I have a TOUGH Question Please Help!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters

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CALLING ALL PHYSICS NERDS!!!! I have a TOUGH Question Please Help!!! 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.

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Hải Đăng 5 months 2021-08-19T12:54:27+00:00 2 Answers 0 views 0

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    0
    2021-08-19T12:55:38+00:00

    Answer:

    a. The electron would have traveled about 6.7\times 10^{-2}\; \rm m before hitting the plate.

    b. The speed of the electron would be approximately 4.3\times 10^{6}\; \rm m \cdot s^{-1} right before it hits the plate. Its vertical velocity at that moment would be approximately 1.7\times 10^{6} \; \rm m \cdot s^{-1}\end{aligned}.

    Explanation:

    Acceleration of this electron

    Look up the following data:

    • Elementary charge: q_e \approx 1.60218\times 10^{-19}\; \rm C (that’s the magnitude of the charge on one electron.)
    • Electron rest mass: m_e \approx 9.10938\times 10^{-31}\; \rm kg.

    Note, that the electric field strength V between the two horizontal plates is constant. Therefore, the electrostatic force that this field exerts on this electron should also be constant. Calculate the size of this force:

    \begin{aligned}F &= q \cdot V\\ & \approx 1.60218\times 10^{-19}\; \rm C \times 4.0\times 10^{2}\; \rm N \cdot C^{-1} \\ & \approx 6.40872 \times 10^{-17}\; \rm N\end{aligned}.

    Assume that the effect of gravity on this electron can be ignored. That electrostatic force would be the only force acting on this electron. Hence, the net force on this electron would simply be approximately 6.40872 \times 10^{-17}\; \rm N.

    Apply Newton’s Second Law of motion to find the acceleration of this electron:

    \begin{aligned}a &= \frac{F(\text{net})}{m} \\ &\approx \frac{6.40872\times 10^{-17}\; \rm N}{9.10938\times 10^{-31}\; \rm kg} \approx 7.03530\times 10^{13}\; \rm m \cdot s^{-2} \end{aligned}.

    Note that the electric field between the two plates is constant. As a result, the electrostatic force on the electron, the net force on this electron, as well as this acceleration will all be constant.

    Also, note that 7.03530\times 10^{13}\; \rm m \cdot s^{-2} is much greater than g \approx 9.81\; \rm m \cdot s^{-2}. That agrees with the assumption that the effect of gravity on this electron is much smaller than that of the electrostatic force.

    Time it takes for the electron to hit the bottom plate

    Note, that the acceleration of this electron is perpendicular to its initial velocity. As a result, that electron would continue to move horizontally at the same speed until it hits the bottom plate. That also allows the electron’s motion in the vertical direction to be considered separately.

    Let u denote the initial vertical velocity of this electron. Since this electron was moving horizontally to begin with, u = 0. Let t denote the time it takes for this electron to run into the bottom plate. Note that the electron was initially at the top plate, which is x = 2.0\; \rm cm = 0.020\; \rm m away from the bottom plate. Apply the SUVAT equation x = (1/2)\, a \, t^2 + u \, t:

    \begin{aligned}\underbrace{0.020}_{x} = \underbrace{\left(\frac{1}{2} \times 7.03530\times 10^{13}\right)}_{(1/2)\, a} \, t^2\end{aligned}.

    Solve for t:

    t \approx 2.38\times 10^{-8}\; \rm s (given that t > 0.)

    Horizontal distance traveled before the electron hits the plate

    Since the horizontal velocity was constant, the displacement in time t can be found as:

    \begin{aligned}\text{horizontal distance} &=\text{horizontal velocity} \times \text{time} \\ &\approx 4.0\times 10^{6}\; \rm m \cdot s^{-1} \times 2.38\times 10^{-8}\; \rm s \\ &\approx 9.5\times 10^{-2}\; \rm m \end{aligned}

    (Rounded to two significant figures.)

    Velocity of the electron right before hitting the plate

    Start by finding the horizontal and vertical velocity at that moment (separately.)

    The acceleration in the vertical direction is constant and has already been found. Therefore:

    \begin{aligned}& (\text{final vertical velocity}) \\ &=(\text{vertical acceleration}) \times (\text{time}) + (\text{initial vertical velocity}) \\ &\approx 7.03530\times 10^{13}\; \rm m \cdot s^{-2} \times 2.38\times 10^{-8}\; \rm s + 0\; \rm m \cdot s^{-1} \\ &\approx 1.67\times 10^{6} \; \rm m \cdot s^{-1}\end{aligned}.

    On the other hand, the horizontal velocity of this electron at that moment should still be 4.0\times 10^{6}\;\rm m \cdot s^{-1}. If the velocity of the electron is to be expressed as a vector, then:

    \displaystyle v = \begin{bmatrix}\text{horizontal velocity}\\ \text{vertical velocity} \end{bmatrix} \approx \begin{bmatrix}4.0 \times 10^{6} \\ 1.67\times 10^{6}\end{bmatrix}\; \rm m \cdot s^{-1}.

    The magnitude of the velocity (also known as speed) of the electron at that moment will be:

    \begin{aligned}\text{speed} &\approx \left(\sqrt{\left(4.0\times 10^6\right)^2 + \left(1.67\times 10^6\right)^2}\right) \; \rm m \cdot s^{-1} \\ &\approx 4.3\times 10^{6}\; \rm m\cdot s^{-1}\end{aligned}.

    (Rounded to two significant figures.)

    0
    2021-08-19T12:56:08+00:00

    Answer:

    1.676m/s

    Explanation:

    here is my work for this question. am I on the right path

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