Question

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?

Answer 1

Answer:

(a) $F = -4.01 * 10^{-15} N$

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

Explanation:

Parameter given:

Electric field, E = $2.5 * 10^4 N/C$

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

Mathematically:

$F = qE$

Electric charge, q, of an electron = $- 1.602 * 10^{-19} C$

$F = -1.602 * 10^{-19} * 2.5 * 10^4\\\\\\F = -4.01 * 10^{-15} N$

(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:

$-ma = -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}$

Mass, m, of an electron = $9.11 * 10^{-31} kg$

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

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