Question

A battery does 2.425 J of work to transfer 0.045 C of charge from the negative to the positive terminal. What is the emf of this battery?

Answer 1

We know that the work done to move a particle in an eletric field is:

`\tau=U.q`

So, if we replace our values, we get

`2.425=U*0.045`

Thus

`U=(2.425)/(0.045)=58.89V`