A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in electrical potential energy of the particle?

(1) 1.07 * 10 19eV

(2) 2.16 * 10 18eV

(3) 1.50 eV

(4) 13.5 eV

The answer:

the relationship between elementary charge, potential difference and electrical potential energy is given by

E = qV

E: lectrical potential energy

q: elementary charge

V: potential difference

but we have e=abs val (q) = 3

so we have E = qV=3ex4.5V = 13.5 eV

the answer is (4) 13.5 eV