Question

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?

Answer 1

Answer:

The work required for the electron is $1.44x10^{-18} J$

Explanation:

The work of the electron is calculated in the equation:

$W= e * V$

$e= 1.6x10^{-19} C$

C= Coulomb = s*A

V=  Volts = $\frac{W}{A}$

W= $\frac{J}{s}$

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