Question

If 27 J of energy is expended in moving 3 C of charge between two points, how much voltage is there between those two points?

Answer 1

the voltage woulde be 12

Answer 2

It has given that, 27 J of energy is expanded in moving 3C of charge between two points.

We have to find the voltage between those two points.

solution : when a particle of charge q moves from point A to point B in a electric field, where potential difference between point A and B is ∆V, then workdone on the charge q is equal to q∆V.

i.e., W = q∆V

as work done is conservative so potential energy of charge q is equal to work done on the charge. i.e., U = W = q∆V

here potential energy = 27 J

charge, q = 3C

so 27 J = 3C × ∆V

⇒∆V = 9 volts.

Therefore voltage between those two points is 9 volts.