A proton (+1.6 × 10−¹⁹ C)moves 10 cm on a path in the direction opposite to a uniform electric field of strength 3.0 N/C.How much work is done on the proton by the electrical field? A) 4.8 × 10−²⁰ J B) 1.6 × 10−²⁰ J C) zero D) −4.8 × 10−²⁰ J

Question

Quizplus · Accepted Answer

The work done by the electric field is given by the formula W = Fd, where W is the work done, F is the force, and d is the displacement. In this case, the force on the proton is given by F = qE, where q is the charge of the proton and E is the electric field strength. Since the proton moves in a direction opposite to the electric field, the force on the proton is also in the opposite direction, i.e., opposite to the displacement. Therefore, the work done by the electric field is negative.

Putting the values, we have: 
q = +1.6 × 10^-19 C (charge of proton)
E = 3.0 N/C (electric field strength)
d = -0.1 m (displacement)

Substituting in the formula, we get:
W = (1.6 × 10^-19 C) x (3.0 N/C) x (-0.1 m) = -4.8 × 10^-20 J

Therefore, the work done on the proton by the electric field is -4.8 × 10^-20 J or approximately -5 × 10^-20 J. The answer is D, -4.8 × 10^-20 J.

A Proton (+1