Question

(a) Deduce the expression for the potential energy of a system of two charges $q_1$ and $q_2$ located $\overrightarrow{r}{}_1$ and $\overrightarrow{r}{}_2$ , respectively, in an external electric field.

(b) Three point charges, + Q + 2Q and – 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-point $A_1$, $B_1$ and $C_1$ ,respectively, find the amount of the work done in shifting the charges to the new locations.

OR

Define electric flux. Write its S.I unit.

State and explain Gauss’s law. Find out the outward flux to a point charge +q placed at the centre of a cube of side ‘a’. Why is it found to be independent of the size and shape of the surface enclosing it? Explain.

Answer 1

The work done in bringing the chare $q_1$ from infinity to $r_2$ can be calculated. Here, the work is done not only against the external field E but also against the field due to $q_1$ Hence, work done on $q_2$ against the external field is $W_2=q_{2}V\left(r_2\right)$

Work done on q against the field due to $q_1$ ,$W_{12}$=$\frac{q_1{q}_2}{4\pi {ϵ}_0{r}_{12}}$

where $r_{12}$ is the distance between $q_1$ and $q_2$

By the principle of superposition for fields, work done on $q_2$ against two fields will add with work done in bringing $q_2$ to $r_2$ , which is given as

$W+W=qV\left(r_2\right)+\frac{q_1{q}_2}{4\pi {ϵ}_0{r}_{12}}$

Thus, the potential energy of the system U = total work done in assembling the configuration U

$W_1+W_2+W{+}_{12}$

$U=qV\left(r_1\right)+qV\left(r_2\right)+\frac{q_1{q}_2}{4\pi {ϵ}_0{r}_{12}}$

(b)

q=+Q

q = +2Q

q = -3Q

r = l (for each side)

Intial potential energy of system

$U_1=\frac{1}{4\pi {ϵ}_{0}1}\left[\right(q_1\times q_2)+(q_2\times q_3)+(q_3\times q_1\left)\right]U_1=\frac{1}{4\pi {ϵ}_{0}1}\left[\right(+Q\times +2Q)+(+2Q\times -3Q)+(-3Q\times +Q)+]U_1=\frac{-7Q_2}{4\pi {ϵ}_{0}1}$

These charges displaced to mid points then final potential energy of system

$U_1=\frac{1}{4\pi {ϵ}_{0}1/2}\left[\right(q_1\times q_2)+(q_2\times q_2)+(q_3\times q_2\left)\right]U_2=\frac{2}{4\pi {ϵ}_{0}1}\left[\right(+Q\times +2Q)+(2Q\times -3Q)+(-3Q\times +Q\left)\right]U_2=\frac{-7Q^2}{4\pi {ϵ}_{0}1}$

$\text{Work done, W}=U_2-U_{1}W=\frac{-7Q^2}{2\pi {ϵ}_{0}1}-\frac{-7Q^2}{4\pi {ϵ}_{0}1}W=\frac{-7Q^2}{\pi {ϵ}_{0}1}[\frac{-1}{2}-\left(\frac{-1}{4}\right)]=\frac{7Q^2}{\pi {ϵ}_{0}1}[\frac{-1}{2}+\frac{1}{4}]W=\frac{-7}{4}\left(\frac{Q^2}{\pi {ϵ}_{0}1}\right)$

OR

Electric flux is the total number of lines of force passing through the unit area of a surface held perpendicularly.

Electric flux$\Delta \varphi$ through an area element $\Delta S$ is given by

$\Delta \varphi =E.\Delta S=E\Delta Scos\theta$

$\theta$is the angle between E and $\Delta S$

The SI unit of electric flux is $N{C}^{-1}{m}^2$

Gauss’s law states that the flux of the electric field through any closed surface S is $1/{\epsilon }_0$ times the total charge enclosed by S.

Let the total flux through a sphere of radius r enclose a point charge q at its centre. Divide the sphere into a small area element as shown in the figure.

The flux through an area element $\Delta S$ is

$\Delta \varphi =E\Delta S=\frac{q}{4\pi {ϵ}_0{r}^2}\hat{r}.\Delta S$

Here, we have used Coulomb’s law for the electric field due to a single charge q.

The unit vector $\hat{r}$ is along the radius vector from the centre to the area element. Because the normal to a sphere at every point is along the radius vector at that point, the area element $\Delta S$ and $\hat{r}$ have the same direction. Therefore,

$\Delta \varphi =\frac{q}{4\pi {ϵ}_0{r}^2}.\Delta S$

Because the magnitude of the unit vector is 1, the total flux through the sphere is obtained by adding the flux through all the different area elements.

$\varphi ={\sum }_{all\Delta S}=\frac{q}{4\pi {ϵ}_0{r}^2}.\Delta S$

Because each area element of the sphere is at the same distance r from the charge,

$\varphi =\frac{q}{4\pi {ϵ}_0{r}^2}{\sum }_{all\Delta S}=\frac{q}{4\pi {ϵ}_0{r}^2}.S$

Now, S the total area of the sphere equals $4\pi r^2$ .Thus

$\varphi =\frac{q}{4\pi {ϵ}_0{r}^2}\times 4\pi r^2=\frac{q}{{ϵ}_0}$

Hence, the above equation is a simple illustration of a general result of electrostatics called Gauss’s law.

Let a cube of side a enclose charge +q at its centre.

Because the electric flux through the square surface is$\varphi =\frac{q}{6{ϵ}_0}$ the square surfaces of cube are six. Hence, according to Gauss’s theorem in electrostatics, the total outward flux due to a charge +q of a cube is $\varphi =6\times \left(\frac{q}{6{ϵ}_0}\right)=\frac{q}{{ϵ}_0}$The result shows that the electric flux passing through a closed surface is proportional to the charge enclosed. In addition, the result reinforces that the flux is independent of the shape and size of the closed surface.