Answer:
Approximately [tex]7.31\; {\rm m}[/tex].
Explanation:
Assume that the only force on the electron is from the electric field. The kinetic energy that the electron gained would be equal to the work that the field did on the electron.
The electron was initially not moving, meaning that the initial kinetic energy of the electron would be [tex]0\; {\rm J}[/tex]. Since the electron eventually gained [tex]2.05 \times 10^{-16}\; {\rm J}[/tex] of kinetic energy, the change in kinetic energy would be:
[tex]\begin{aligned}(\text{change in KE}) &= (\text{current KE}) - (\text{initial KE}) \\ &= (2.05 \times 10^{-16}\; {\rm J}) - (0\; {\rm J}) \\ &= 2.05 \times 10^{-16}\; {\rm J}\end{aligned}[/tex].
The amount of work that the field did on the electron would be equal to the gain in kinetic energy: [tex]2.05 \times 10^{-16}\; {\rm J}[/tex].
Since the force on the electron is constant, dividing work by force would give the displacement in the direction of the force. The electrostatic force on this electron would be equal to the product of the charge on the electron and the electric field.
The magnitude of the electrostatic charge on an electron is known as the elementary charge: approximately [tex]1.602 \times 10^{-19}\; {\rm C}[/tex]. Given that the electric field strength is [tex]175\; {\rm N \cdot C^{-1}}[/tex], the magnitude of the electrostatic force on this electron would be:
[tex]\begin{aligned}(175\; {\rm N\cdot C^{-1}})\, (1.602 \times 10^{-19}\; {\rm C}) \approx 2.8035 \times 10^{-17}\; {\rm N}\end{aligned}[/tex].
Divide work by force to find the distance travelled:
[tex]\displaystyle \frac{2.05 \times 10^{-16}\; {\rm J}}{2.8035\times 10^{-17}\; {\rm N}} \approx 7.31\; {\rm m}[/tex].
(Note that [tex]1\; {\rm J} = 1\; {\rm N\cdot m^{-1}}[/tex].)
