What is the potential in the equatorial plane of a dipole having dipole moment p?
OR
(i) Eight dipoles of charges of magnitude e each are placed inside a cube. What will be the total electric flux coming out of the cube? 

(ii) Two parallel infinite line charges $+\mathrm{\lambda }$ and $-\mathrm{\lambda }$ are placed with a separation distance R in free space. The net electric field exactly mid-way between the two line charges is An infinite line charge produces a field of 18 x 10⁴ N/C at 0.02 m. What is the linear charge density?
 

$$\begin{gathered} \mathrm{The} \mathrm{potential} \mathrm{due} \mathrm{to} \mathrm{an} \mathrm{electric} \mathrm{dipole} \mathrm{is} \\ \mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{p}.\mathrm{r}}{{\mathrm{r}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{\mathrm{pr} \cos \mathrm{\theta }}{{\mathrm{r}}^2} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{\mathrm{pr} \cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{{\mathrm{r}}^2}\left[\therefore \mathrm{\theta }=\frac{\mathrm{\pi }}{2}\right] \\ =0 \end{gathered}$$

A parallel plate air capacitor has a capacitance 18 µF. If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes 72 µF. The dielectric constant of the medium is 

Three capacitors 3µF, 6µF and 6µF are connected in series to a source of 120 V. The potential difference (in volt) across the 3µF capacitor will be

$$\begin{gathered} \therefore {\mathrm{C}}_0=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}=18 \dots \left(\mathrm{i}\right) \\ \mathrm{and} \mathrm{C}=\frac{{\mathrm{K\epsilon }}_0\mathrm{A}}{3\mathrm{d}}=72 \dots \left(\mathrm{ii}\right) \\ \mathrm{On} \mathrm{dividing} \mathrm{Eq}. \left(\mathrm{ii}\right) \mathrm{by} \mathrm{Eq}. \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \frac{\mathrm{K}}{3}=\frac{72}{18}=4 \\ \therefore \mathrm{Dielectric} \mathrm{constant}, \mathrm{K}=12 \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{combination} \mathrm{of} \mathrm{three} \mathrm{capacitors} \mathrm{in} \mathrm{series}, \\ \frac{1}{\mathrm{C}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}+\frac{1}{{\mathrm{C}}_3} \\ =\frac{1}{3}+\frac{1}{6}+\frac{1}{6} \\ \Rightarrow \mathrm{C}=\frac{6}{4}=1.5\mathrm{\mu F} \\ \mathrm{The} \mathrm{charge} \mathrm{of} \mathrm{this} \mathrm{circuit}, \\ \mathrm{q}=\mathrm{CV}=1.5\times 120=180 \mathrm{\mu C} \\ \mathrm{The} \mathrm{potential} \mathrm{difference} \mathrm{across} 3\mathrm{\mu F}, \\ \mathrm{q}=\mathrm{CV} \\ \Rightarrow \mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}=\frac{180}{3}=60 \mathrm{V} \end{gathered}$$

(i) Net charge of one dipole = - e +e =  0 

Net charges of 8 dipoles = 8 x 0 = 0 

Net charge inside cube, q = 0 

By Gauss’s law, total flux emerging from surface,

$\mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}=\frac{0}{{\mathrm{\epsilon }}_0}=0$

(ii) 

$$\begin{gathered} \mathrm{According} \mathrm{to} \mathrm{Gauss}\text{'}\mathrm{s} \mathrm{theorem}, \mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0} \\ \mathrm{or} \mathrm{E}.2\mathrm{\pi R}\times \mathrm{l}=\frac{\mathrm{\lambda }}{{\mathrm{\epsilon }}_0} \\ \mathrm{Electric} \mathrm{field} \mathrm{in} \mathrm{mid}-\mathrm{way} \mathrm{due} \mathrm{to} \mathrm{both} \mathrm{lines}, \\ \mathrm{E}=\frac{1}{2\mathrm{\pi }}\frac{\mathrm{\lambda }}{{\mathrm{\epsilon }}_0(\mathrm{R}/2)}\times 2 \\ \Rightarrow \mathrm{E}=\frac{2\mathrm{\lambda }}{{\mathrm{\pi \epsilon }}_0\mathrm{R}} \\ \mathrm{Linear} \mathrm{charge} \mathrm{density}, \\ \mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}=\frac{2\mathrm{\lambda }}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}} \\ \mathrm{or} \mathrm{\lambda }=\frac{\mathrm{E}\times 4{\mathrm{\pi \epsilon }}_0\mathrm{r}}{2} \\ =18\times {10}^4\times \frac{1}{9\times {10}^9}\times \frac{0.02}{2} \\ =2\times {10}^{-7}\mathrm{C}/\mathrm{m} \end{gathered}$$