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

$U=\frac{2.425}{0.045}=58.89V$