Electrons in a particle beam each have a kinetic energy of 4.0 × 10−¹⁷ J.What is the magnitude of the electric field that will stop these electrons in a distance of 0.3 m? (e = 1.6 × 10−¹⁹ C) A)1.9E+3 N/C B)7.5E+3 N/C C)8.3E+2 N/C D)1.3E+3 N/C E)6.3E+3 N/C

Question

Quizplus · Accepted Answer

The work done by an electric field on a charge is given by the equation:

W = qΔV

where q is the charge, ΔV is the potential difference, and W is the work done.

Since the electrons are being stopped, their final kinetic energy is zero. Therefore, the work done by the electric field is equal to the initial kinetic energy of the electrons:

W = ΔK = 4.0 × 10^-11 J

The potential difference ΔV is related to the electric field E and the distance d:

ΔV = Ed

Therefore,

E = ΔV / d

Substituting the given values, we get:

E = (4.0 × 10^-11 J) / (1.6 × 10^-19 C) / (0.3 m)

E = 8.3 × 10^2 N/C

Therefore, the answer is C.

Electrons in a Particle Beam Each Have a Kinetic Energy