Question

What is the change in electric potential difference when +30.0 J of work is done on a charge of +0.010 C?

-300 V , +300 V , +3000 V , or -3000 V

Answer 1

positive charge will move towards higher potential, so the requested differential of potential will be positive, furthermore, the subsequent relationship, holds:
\[\Large \Delta V = \frac{L}{q}\]

Answer 2

ok:)

Answer 3

namely:
\[\Large \Delta V = \frac{L}{q} = \frac{{30}}{{{{10}^{ - 2}}}}\]

Answer 4

i get this?
3000?

so my answer is +3000 V ?

Answer 5

yes! that's right!

Answer 6

yay! :) thanks!!

Answer 7

:)