Question

How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of electrons from an initial point where the electric potential is 6.00 V to a point where the potential is -5.00 V

Answer 1

`1.06* 10^6 J`

Step-by-step explanation:

We are given that

`V_1=6 V`

`V_2=-5 V`

Charge on 1 electron, 1 e=
`-1.6* 10^(-19) C`

Avogadro's number,
`N_A=6.02* 10^(23)`

`q=ne`

`q=6.02* 10^(23)* 1.6* 10^(-19)=-9.632* 10^4 C`

`W=q\Delta V`

`W=q(V_2-V_1)`

`W=-9.632* 10^4* (-5-6)=1.06* 10^6 J`

Hence, the work done=
`1.06* 10^6 J`