Question

A voltage of 18 V is applied across the ends of a piece of copper wire 8 em long. The mass of an electron is 9.11 x 10-31 kg and its charge Is 1.6 × 10-19 C. What is the magnitude of the electron's acceleration?

Answer 1

`3.96* 10^(16)` m/s²

Step-by-step explanation:

V = Potential difference between the ends of copper wire = 18 Volts

d = distance between the ends of the wire = 8 cm = 0.08 m

E = Electric field between the ends

Using the equation

V = E d

18 = E (0.08)

E = 225 N/C

m = mass of electron = 9.1 x 10⁻³¹ kg

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

Magnitude of acceleration of electron is given as

`a = (qE)/(m)`

`a = ((1.6* 10^(-19))(225))/(9.1* 10^(-31))`

`a = 3.96* 10^(16)` m/s²