Question

Question: How much work is required to set up the arrangement of Fig. 24-52 if, q =2.30 pC, a = 64.0 cm and the particles are initially infinitely far apart and at rest?

Short answer

Answer:

The work required to set up the arrangement is $-0.192\text{pJ}$.

Step-by-step solution

The given data

1. Charge of the particle,$q=2.30\text{pC}$

The length of the square, a =0.64m

Understanding the concept of the work done

Using the concept of the electric potential of all the points charges present on the square, we can get the difference between the electric potential at the center of the square. Thus, this value multiplied by the charge on which external work is done determines the work done.

Formulae:

Expression for the potential energy at a point due to a point chargesis given:

$U=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1{q}_2}{r}\right)$ (i)

Here, q is the charge and r is the distance between the charges.

Expression for the work done to move the given charge from the infinity to the given point is given by, $W=\Delta U$ (ii)

Calculation of the work done

Length of the diagonal of the square is given as:

$$\begin{gathered} d=\sqrt{a^2+a^2} \\ =\sqrt{2}a \end{gathered}$$

Initial potential energy, $U_i=0\text{J}$(since V = 0 at infinity)

Thus, the final potential energy of the given system due to all the present charges is given using equation (i) as follows:

$$\begin{gathered} U_f=U_{12}+U_{23}+U_{34}+U_{14}+U_{13}+U_{24} \\ =\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1{q}_2}{a}+\frac{q_2{q}_3}{a}+\frac{q_3{q}_4}{a}+\frac{q_1{q}_4}{a}+\frac{q_1{q}_3}{a}+\frac{q_2{q}_4}{a}\right] \\ =\frac{1}{4\pi {\epsilon }_0}\left[\frac{\left(-q\right)\left(q\right)}{a}+\frac{\left(q\right)\left(-q\right)}{a}+\frac{\left(-q\right)\left(q\right)}{a}+\frac{\left(-q\right)\left(q\right)}{a}+\frac{\left(-q\right)\left(-q\right)}{\sqrt{2}a}+\frac{\left(q\right)\left(q\right)}{\sqrt{2}a}\right] \\ =\frac{q^2}{4\pi {\epsilon }_0}\left[\frac{-1}{a}-\frac{1}{a}-\frac{1}{a}-\frac{1}{a}+\frac{1}{\sqrt{2}a}+\frac{1}{\sqrt{2}a}\right] \\ =\frac{2q^2}{4\pi {\epsilon }_0}\left(\frac{1}{\sqrt{2}}-2\right) \\ =\frac{2\left(9\times {10}^9\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right){(2.3\times {10}^{-12}\text{C})}^2}{\left(0.64\text{m}\right)}\left(\frac{1}{\sqrt{2}}-2\right) \\ =-1.92\times {10}^{-13}\text{J} \end{gathered}$$

Now, the work using the above data in equation (ii) is given as follows:

$$\begin{gathered} W=\left(-1.92\times {10}^{-13}\text{J}\right)-0 \\ =-1.92\times {10}^{-13}\text{J} \\ \text{=}-0.192\text{pJ} \end{gathered}$$

Hence, the value of the work is $-0.192\text{pJ}$.