Suggested languages for you:

Americas

Europe

66P

Expert-verifiedFound in: Page 473

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**Question66:(III) In a given CRT, electrons are accelerated horizontally by 9.0 kV. They then pass through a uniform electric field E for a distance of 2.8 cm, which deflects them upward so they travel 22 cm to the top of the screen, 11 cm above the center. Estimate the value of E.**

The value of the electric field is estimated to be \(3.7 \times {10^5}\;{\rm{V/m}}\).

A cathode ray tube (CRT) was traditionally used in televisions and computer monitors. **In the cathode ray tube, a high voltage anode initially accelerates the electrons produced by the cathode horizontally. These electrons are then passed through the magnetic coils or electric deflection plates, which deflect these electrons onto the screen.**

Mass of the electron is, \({m_1} = 9.11 \times {10^{ - 31}}\;{\rm{kg}}\)

Charge on the electron is, \(q = e = 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

The accelerating potential between the cathode and anode plate is, \(V = 9.0\;{\rm{kV}} = 9.{\rm{0}} \times {\rm{1}}{{\rm{0}}^3}\;{\rm{V}}\)

Distance through which electron passes in the electric field *E* is, \(d = 2.8\;{\rm{cm}} = 2.{\rm{8}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}\)

After emerging out of the deflecting plates:

Distance traveled by the electron to the top of the screen is, \({\rm{AB}} = 22\;{\rm{cm}} = 2{\rm{2}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}\)

Vertical distance traveled by the electron from the center is, \({\rm{AC}} = y = 11\;{\rm{cm}} = 11 \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}\)

Let the horizontal distance traveled by the electron to reach the center of the screen is BC = *x*.

The path of electrons while passing through CRT is shown in the figure below:

When the electrons are accelerated horizontally by potential *V*, then they acquire the horizontal velocity \({v_{\rm{x}}}\). Thus, you can say that the initial potential energy of the electrons gets converted into kinetic energy.

\(\begin{array}{l}PE = KE\\eV = \frac{1}{2}mv_{\rm{x}}^{\rm{2}}\end{array}\)

\({v_{\rm{x}}} = \sqrt {\frac{{2eV}}{m}} \) … (i)

When electrons pass through the uniform electric field, they get deflected upwards by the electric force and thus acquire velocity \({v_y}\)and acceleration *a*. Thus, the electric force on the electrons acting in the upward direction is:

\(\begin{array}{c}F = qE\\ma = eE\\a = \frac{{eE}}{m}\end{array}\)

Since an electron travels the horizontal distance *d* in electric field with velocity\({v_{\rm{x}}}\), thus time for which an electron remains in the field is:

\(t = \frac{d}{{{v_{\rm{x}}}}}\)

Using kinematics equation of motion, vertical velocity acquired by the electrons on passing through the electric field is:

\(\begin{array}{c}{v_y} = 0 + at\\ = \left( {\frac{{eE}}{m}} \right)\left( {\frac{d}{{{v_x}}}} \right)\end{array}\)...... (ii)

After leaving the electric field *E*, the electrons travel along path BA and reach the top of the screen.

In \(\Delta {\rm{ABC}}\),

\(\begin{array}{c}\sin \theta = \frac{{AC}}{{AB}} = \frac{{11\;{\rm{cm}}}}{{22\;{\rm{cm}}}}\\\theta = 30^\circ \end{array}\)

Thus, the electrons move at an angle of \(30^\circ \) after leaving the electric field. So, the horizontal distance traveled by the electrons is:

\(\begin{array}{c}x = AC\cos \theta \\ = \left( {22 \times {{10}^{ - 2}}\;{\rm{m}}} \right)\cos 30^\circ \\ = 19.05 \times {10^{ - 2}}\;{\rm{m}}\end{array}\)

If an electron takes time \(t'\)to reach the screen after leaving the electric field, then the ratio of vertical and horizontal distances traveled by the electron can be written as:

\(\frac{y}{x} = \frac{{{v_{\rm{y}}}t'}}{{{v_x}t'}} = \frac{{{v_{\rm{y}}}}}{{{v_x}}}\)

On substituting the value of \({v_y}\)from equation (i) in the above expression, you will get:

\(\begin{array}{c}\frac{y}{x} = \frac{{\left( {\frac{{eE}}{m}} \right)\left( {\frac{d}{{{v_{\rm{x}}}}}} \right)}}{{{v_{\rm{x}}}}}\\ = \frac{{eEd}}{{mv_{\rm{x}}^{\rm{2}}}}\\E = \frac{y}{x} \times \frac{{mv_{\rm{x}}^{\rm{2}}}}{{ed}}\end{array}\)

Using (i), you will get:

\(\begin{array}{c}E = \frac{y}{x} \times \frac{m}{{ed}}\left( {\frac{{2eV}}{m}} \right)\\ = \frac{y}{x} \times \frac{{2V}}{d}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}E = \left( {\frac{{11 \times {{10}^{ - 2}}\;{\rm{m}}}}{{19.05 \times {{10}^{ - 2}}\;{\rm{m}}}}} \right) \times \frac{{2\left( {9.{\rm{0}} \times {\rm{1}}{{\rm{0}}^3}\;{\rm{V}}} \right)}}{{\left( {2.{\rm{8}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}} \right)}}\\ = 3.7 \times {10^5}\;{\rm{V/m}}\end{array}\)

Thus, the value of *E* is estimated to be \(3.7 \times {10^5}\;{\rm{V/m}}\).

94% of StudySmarter users get better grades.

Sign up for free