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

(a)
`F = -4.01 * 10^(-15) N`

(b)
`a = 4.40 * 10^(15) m/s^2`

Step-by-step 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 = (-4.01 * 10^(-15))/(-m)`

Mass, m, of an electron =
`9.11 * 10^(-31) kg`

=>
`a = (-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`