contestada

An electron is carried from the positive terminal to the negative terminal of a 9 v battery. How much work is required in carrying this electron?

Respuesta :

Answer:

The work required for the electron is [tex]1.44x10^{-18} J[/tex]

Explanation:

The work of the electron is calculated in the equation:

[tex]W= e * V[/tex]

[tex]e= 1.6x10^{-19} C[/tex]

C= Coulomb = s*A

V=  Volts = [tex]\frac{W}{A}[/tex]

W= [tex]\frac{J}{s}[/tex]

[tex]W= 1.6 x10^{-19}C *9 V\\W=1.44x10^{-18} C*V\\ W=1.44x10^{-18} s*A*\frac{W}{A} \\W=1.44x10^{-18} W*s\\W=1.44x10^{-18} \frac{J}{s}*s \\W=1.44x10^{-18} J\\[/tex]