Question

The electron beam emitted by an electron gun is controlled (steered) with two sets of parallel conducting plates: a horizontal set to control the vertical motion of the beam, and a vertical set to control the horizontal motion of the beam. The beam is emitted with an initial velocity of \(2.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s} .\) The width of the plates is \(d=5.00 \mathrm{~cm},\) the separation between the plates is \(D=4.00 \mathrm{~cm},\) and the distance between the edge of the plates and a target screen is \(L=40.0 \mathrm{~cm} .\) In the absence of any applied voltage, the electron beam hits the origin of the \(x y\) -coordinate system on the observation screen. What voltages need to be applied to the two sets of plates for the electron beam to hit a target placed on the observation screen at coordinates \((x, y)=(0 \mathrm{~cm}, 8.00 \mathrm{~cm}) ?\)

Short answer

Answer: To find the applied voltage needed for the electron beam to hit the target, we can use the formula derived in step 5: \(V = \frac{2m_eyD(v_0)^2}{ed^2}\), and plug in the given values for \(e, m_e, y, D, v_0,\) and \(d\). This will give us the required applied voltage.

Step-by-step solution

Identify variables and constants

First, let's list down all the given variables and constants in the problem. - Initial velocity of the electron beam, \(v_0 = 2.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) - Width of plates, \(d = 5.00 \mathrm{~cm} = 0.05 \mathrm{~m}\) - Separation between plates, \(D = 4.00 \mathrm{~cm} = 0.04 \mathrm{~m}\) - Distance between the edge of plates and target screen, \(L = 40.0 \mathrm{~cm} = 0.4 \mathrm{~m}\) - Target coordinates, \((x, y) = (0, 8.00 \mathrm{~cm}) = (0, 0.08 \mathrm{~m})\) - We will also use the charge of an electron, \(e = 1.60 \times 10^{-19} \mathrm{C}\), and its mass, \(m_e = 9.11 \times 10^{-31} \mathrm{kg}\)

Find the time taken to move horizontally

The electron beam moves horizontally through the plates with an initial velocity \(v_0\). As it is not affected by any force horizontally, it will have a constant horizontal velocity throughout. Calculate the time it takes for the electron beam to pass through the horizontal plates using the formula: \(t = \frac{d}{v_0}\) where \(d\) is the width of the plates and \(v_0\) is the initial velocity of the electron beam.

Calculate acceleration due to vertical electric field

The force experienced by the electron in the vertical electric field is given by \(F_e = eE\), where \(e\) is the charge of the electron and \(E\) is the electric field. The electric field between parallel plates can be calculated as \(E = \frac{V}{D}\), where \(V\) is the voltage across the plates and \(D\) is the separation between the plates. By Newton's second law, the vertical acceleration of the electron is given by \(a_y = \frac{F_e}{m_e} = \frac{eE}{m_e}\) Thus, the vertical acceleration can also be expressed as \(a_y = \frac{eV}{m_eD}\).

Calculate vertical displacement of the electron

By calculating the vertical displacement of the electron, we solve the problem when the electron beam reaches the target point. The equation of vertical displacement is given by \(y = v_{0_y}t + \frac{1}{2}a_y t^2\) Since there is no initial vertical velocity, \(v_{0_y} = 0\), the equation becomes \(y = \frac{1}{2}a_y t^2\) Replacing \(a_y\) and \(t\) from steps 2 and 3, we get \(y = \frac{1}{2}\left(\frac{eV}{m_eD}\right)\left(\frac{d}{v_0}\right)^2\)

Calculate the applied voltage

For the electron beam to hit the target \((x, y) = (0, 0.08 \mathrm{~m})\), we need to find the applied voltage \(V\). Rerranging the equation from Step 4: \(V = \frac{2m_eyD(v_0)^2}{ed^2}\) Now plug in the values for \(e, m_e, y, D, v_0,\) and \(d\) and solve for \(V\). To recap, the integral steps to solving this problem involve: 1. Identifying variables and constants given in the problem 2. Calculating the time taken for an electron to travel through horizontal plates 3. Determining the acceleration due to the vertical electric field 4. Calculating the vertical displacement of the electron 5. Calculating the applied voltage to both sets of plates, in order to guide the electron beam to the target