Question

A proton (+1.6 × 10−19 C) moves 20 cm on a path in the direction of a uniform electric field of strength 2.4 N/C. How much work is done on the proton by the electrical field?

Answer 1

W=Fd

F=Eq

W=Eqd

E = 2.4 N/C

q = 1.6×10⁻¹⁹ C

d = 20 cm = 0.2 m

W = 2.4(1.6×10⁻¹⁹)(0.2)

W = 7.68×10⁻²⁰ J