Question

An electron traveling through a curving wire in an electric circuit experiences a constant force of $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathbf{N}$, always in the direction of its motion through the wire. How much work is done on the electron by this force as it travels through 0.5 m of the wire?

Short answer

The work done on the electron is$2.5\times {10}^{-19}\mathrm{J}$

Step-by-step solution

Identification of given Data

The given data is listed below as,

• The force experience by the electron is,$F=5\times {10}^{-19}\mathrm{N}$
• Distance moved by the electron is, $x=0.5\mathrm{m}$

Definition of the work done

The work done as a result of the force is given by the following expression,

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}\mathbf{.}}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{F}\mathbf{}\mathbf{.}\mathbf{xcos}\mathbf{}\mathbf{\theta } \\ \mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

Here, F is the external force and x is displacement and $\theta$is the angle between force and distance moved.

Step 2: Determination of the work done on the electron

The angle between force and distance moved is .

Substitute all the values in equation (1).

$$\begin{gathered} W=\left(5\times {10}^{-19}\mathrm{N}\right)\left(0.5\mathrm{m}\right)\cos 180° \\ =-\left(5\times {10}^{-19}\mathrm{N}\right)\left(0.5\mathrm{m}\right) \\ =-2.5\times {10}^{-19}\mathrm{N}.\mathrm{m}\times \frac{1\mathrm{N}.\mathrm{m}}{1\mathrm{J}} \\ =-2.5\times {10}^{-19}\mathrm{J} \end{gathered}$$

The work is negative this indicates that the force is applied in the opposite direction of the displacement.

Thus, the work done on the electron is $2.5\times {10}^{-19}\mathrm{J}$.