A proton is moved so that its electric potential energy increases from 2.0 × 10-18 J to 6.0 × 10-18 J. The magnitude of the charge on a proton is 1.602 × 10-19 C. The electric potential difference through which the proton moved, to the nearest whole number, is
V.

radiation

explanation:

heat is transmitted by radiation

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j