Question

Question: In Fig. 24-54, how much work must we do to bring a particle, of charge$\mathbf{Q}\mathbf{=}\mathbf{+}\mathbf{16}\mathbf{e}$and initially at rest, along the dashed line from infinity to the indicated point near two fixed particles of charges${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{4}\mathit{E}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{q}}_{\mathbf{1}}\mathbf{/}\mathbf{2}$? Distance,$\mathbf{}\mathit{d}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{\text{cm,}}{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{43}\mathbf{°}\mathit{a}\mathit{n}\mathit{d}{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{60}\mathbf{°}$

Short answer

Answer:

The work that must be done to bring the particle is 0J.

Step-by-step solution

The given data

The given values of charges and distance from the figure: ,${\text{Q=+16e,q}}_1=+4e,q_2=-2e,d=140cm,{\theta }_1={43}^{o}and{\theta }_2={60}^o$

Understanding the concept of the work done

We know that the work done by a body is given by the difference between the potential energies of the system. Thus, for initial potential energy as zero, as the particle brought from infinity, we can say the work is equal to the final potential energy.

Formula:

The potential energy of the system, $U=\frac{k{q}_1{q}_2}{d}$

Calculation of the work done

After reading the question, and from the given figure,

$$\begin{gathered} W=\Delta U \\ =U_f-U_i \\ =U_f-0 \\ =U_f \end{gathered}$$

Thus, the total required work done is given using equation (i) as follows:

$$\begin{gathered} W=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_{1}Q}{2d}+\frac{q_{2}Q}{d}\right] \\ =9.0\times {10}^9\left[\frac{4\times 16}{0.028}-\frac{2\times 16}{0.014}\right]\times {\left(1.6\times {10}^{-19}\right)}^2 \\ =0\text{J} \end{gathered}$$

Hence, the value of the work done is 0J.