Question

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distance of 1.2 cm. What is the electric field strength?

Answer 1

`E = 2.84 * 10^5 N/C`

Step-by-step explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

`v^2 = u^2 + 2as`

`(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^(15)= 0.024a\\\\a = 1.2 * 10^(15) / 0.024\\\\a = 5 * 10^(16) m/s^2`

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

`E = (9.11 * 10^(-31) * 5 * 10^(16))/(1.602 * 10^(-19)) \\\\E = 2.84 * 10^5 N/C`