Question

A charge of 3.947 nC is moved from a position on the y axis of 8.216 cm to a position on the x axis of 2.848 cm while there is a charge 35.083 nC located at the origin. How much work in micro-Joules did it take to move the charge?

Answer 1

Given:

The charge on the Y-axis is,

$\begin{gathered} q_1=3.947\text{ nC} \\ =3.947\times10^{-9}\text{ C} \end{gathered}$

The initial position of the charge is,

$\begin{gathered} y=8.216\text{ cm} \\ =0.08216\text{ m} \end{gathered}$

The final position of the charge is,

$\begin{gathered} x=2.848\text{ cm} \\ =0.02848\text{ m} \end{gathered}$

The charge at the origin is,

$\begin{gathered} q_2=35.083\text{ nC} \\ =35.083\times10^{-9}\text{ C} \end{gathered}$

To find:

The work to move the charge

Explanation:

The potential at the first point is,

$\begin{gathered} V_1=\frac{kq_2}{y} \\ k=9\times10^9\text{ N.m}^2.C^{-2} \end{gathered}$

The potential at the final point is,

$V_2=\frac{kq_2}{x}$

The work in this process is,

$\begin{gathered} W=q_1(V_2-V_1) \\ =kq_1q_2(\frac{1}{x}-\frac{1}{y}) \\ =9\times10^9\times3.947\times10^{-9}\times35.083\times10^{-9}(\frac{1}{0.02848}-\frac{1}{0.08216}) \\ =2.86\times10^{-5} \\ =28.6\times10^{-6}\text{ J} \\ =28.6\text{ }\mu J \end{gathered}$

Hence, the work is,

$28.6\text{ }\mu J$