Question

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) Group of answer choices

Answer 1

Explanation:

Relation between work and change in kinetic energy is as follows.

                 $W_{net} = \Delta K$

Also,   $\Delta K = K_{initial} - K_{final}$

                        = $(0 - 4.0 \times 10^{-17})$ J

                        = $-4.0 \times 10^{-17}$ J

Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.

                w = -Fd

       $-4.0 \times 10^{-17} J = -F \times 0.3 m$

                F = $1.33 \times 10^{-16}$  

Therefore, relation between electric field and force is as follows.

              E = $\frac{F}{q}$

                 = $\frac{1.33 \times 10^{-16}}{1.60 \times 10^{-19} C}$

                 = $0.831 \times 10^{3}$ C

Thus, we can conclude that magnitude of the electric field that will stop these electrons in a distance of 0.3 m is $0.831 \times 10^{3}$ C.