The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determine the force (use dimensional analysis!) and resulting acceleration of an electron?

Respuesta :

Answer:

(a) [tex]F = -4.01 * 10^{-15} N[/tex]

(b) [tex]a = 4.40 * 10^{15} m/s^2[/tex]

Explanation:

Parameter given:

Electric field, E = [tex]2.5 * 10^4 N/C[/tex]

(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.

Mathematically:

[tex]F = qE[/tex]

Electric charge, q, of an electron = [tex]- 1.602 * 10^{-19} C[/tex]

[tex]F = -1.602 * 10^{-19} * 2.5 * 10^4\\\\\\F = -4.01 * 10^{-15} N[/tex]

(b) This electrostatic force causes the electron to accelerate with an equivalent force:

F = -ma

where m = mass of an electron

a = acceleration of electron

(Note: the force is negative cos the direction of the force is opposite the direction of the electron)

Therefore:

[tex]-ma = -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}[/tex]

Mass, m, of an electron = [tex]9.11 * 10^{-31} kg[/tex]

=> [tex]a = \frac{-4.01 * 10^{-15}}{-9.11 * 10^{-31}}\\\\\\a = 4.40 * 10^{15} m/s^2[/tex]

The acceleration of the electron is [tex]4.40 * 10^{15} m/s^2[/tex]