Question

What is the emf of a battery that increases the electric potential energy of 0.060 CC of charge by 0.70 JJ as it moves it from the negative to the positive terminal?

Answer 1

Answer:

emf = 11.667 V

Explanation:

Given: charge q = 0.060 C, electric potential energy E =0.70 J,

Solution :

by definition 1 volt = 1 joule per coulomb

so Voltage = emf = E/C

emf =  0.70 J / 0.060 C

emf = 11.667 V

Answer 2

Answer:

11.67 V.

Explanation:

Workdone by a charge, W = q × v

Where,

W = 0.7 J

q = 0.06 C

v = W/q

= 0.7/0.06

= 11.67 V

Emf = 11.67 V