Question

Figure 18.57 shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is 3.00×10⁶ m/s , and the horizontal distance it travels in the uniform field is 4.00 cm.

(a) What is its vertical deflection?

(b) What is the vertical component of its final velocity?

(c) At what angle does it exit? Neglect any edge effects.

Answer 1

(a)

Expression for the horizontal distance \(L\) covered by the electron is,

\(L=v_{0 x} t\)

Here, \(v_{0 x}\) is the horizontal component of the velocity of the electron and \(t\) is the time taken to cover the distance.

Rearrange the above equation for \(t\).

\(t=\frac{L}{v_{0 x}}\)

Expression for the magnitude of electrostatic force \(F\) acting on the electron is,

$$ F=q E \ldots \ldots \text { (1) } $$

Here, \(q\) is a charge of an electron and \(E\) is the electric field between the plates.

Expression for Newton's second law of motion is,

\(F=m a_{y} \ldots \ldots\) (2)

Here, \(m\) is the mass of the electron and \(a_{y}\) is the acceleration in the vertical direction.

Equate equation (1) and equation (2).

\(\begin{aligned} m a_{y} &=q E \\ a_{y} &=\frac{q E}{m} \end{aligned}\)

Expression for the vertical distance \(y\) covered by the electron is,

\(y=v_{0 y} t+\frac{1}{2} a_{y} t^{2}\)

Here, \(v_{0 y}\) is the initial velocity in the vertical direction.

Substitute \(d\) for \(y, 0\) for \(v_{0 y}, \frac{q E}{m}\) for \(a_{y}\), and \(\frac{L}{v_{0 x}}\) for \(t\) to find an expression for \(d\).

\(\begin{aligned} d &=(0)\left(\frac{L}{v_{0 x}}\right)+\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{L}{v_{0 x}}\right)^{2} \\ &=\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{L}{v_{0 x}}\right)^{2} \end{aligned}\)

Substitute \(-1.60 \times 10^{-19} \mathrm{C}\) for \(q, 4.00 \mathrm{~cm}\) for \(L,-100 \mathrm{~N} / \mathrm{C}\) for \(E, 9.11 \times 10^{-31} \mathrm{~kg}\) for \(m\),

and \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) for \(v_{0 x}\) to find \(d\).

\(\begin{aligned} d=& \frac{\left(-1.60 \times 10^{-19} \mathrm{C}\right)(-100 \mathrm{~N} / \mathrm{C})\left(4 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right)\right)^{2}}{2\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)^{2}} \\=& 1.56 \times 10^{-3} \mathrm{~m} \end{aligned}\)

Therefore, the vertical deflection is \(1.56 \times 10^{-3} \mathrm{~m}\).

(b)

Expression for the vertical component of final velocity is,

\(v_{y}=v_{0 y}+a_{y} t\)

Substitute 0 for \(v_{0 y}, \frac{q E}{m}\) for \(a_{y}\) and \(\frac{L}{v_{0 x}}\) for \(t\) to find \(v_{y}\).

\(v_{y}=0+\left(\frac{q E}{m}\right)\left(\frac{L}{v_{0 x}}\right)\)

\(v_{y}=\frac{q E L}{m v_{0 x}}\)

Substitute \(-1.60 \times 10^{-19} \mathrm{C}\) for \(q,-100 \mathrm{~N} / \mathrm{C}\) for \(E, 3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) for \(v_{0 x}, 9.11 \times 10^{-31} \mathrm{~kg}\) for

\(m\) and \(4.00 \mathrm{~cm}\) for \(L\) to find \(v_{y}\).

\(\begin{aligned} v_{y} &=\frac{\left(-1.60 \times 10^{-19} \mathrm{C}\right)(-100 \mathrm{~N} / \mathrm{C})\left((4.00 \mathrm{~cm})\left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right)\right)}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)} \\ &=2.342 \times 10^{5} \mathrm{~m} / \mathrm{s} \end{aligned}\)

Therefore, the vertical component of the final velocity is approximately \(2.34 \times 10^{5} \mathrm{~m} / \mathrm{s}\).

(c)

Expression for the angle with which electron exits is,

\(\tan \theta=\frac{v_{y}}{v_{0 x}}\)

Here, \(\theta\) is the angle with which the electron exits.

Substitute \(2.34 \times 10^{5} \mathrm{~m} / \mathrm{s}\) for \(v_{y}\) and \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) for \(v_{0 x}\) to find \(\theta\).

\(\begin{aligned} \tan \theta &=\frac{2.34 \times 10^{5} \mathrm{~m} / \mathrm{s}}{3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}} \\ \theta &=4.46^{\circ} \end{aligned}\)

Therefore, the angle with which the electron exits is \(4.46^{\circ}\).

The vertical deflection is \(1.56 \times 10^{-3} \mathrm{~m}\).

The vertical component of the final velocity is approximately \(2.34 \times 10^{5} \mathrm{~m} / \mathrm{s}\).

The angle with which the electron exits is \(4.46^{\circ}\).