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

Answer:

$3.96\times 10^{16}$ m/s²

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

$a = \frac{qE}{m}$

$a = \frac{(1.6\times 10^{-19})(225)}{9.1\times 10^{-31}}$

$a = 3.96\times 10^{16}$ m/s²