Match the following descriptions with each of the diagrams given in figure. Fields are near the interface, but on opposite sides of the boundary. Vectors are drawn to scale.
(a) Medium 1 and medium 2 are dielectrics with \varepsilon_{1}>\varepsilon_{1}.
(b) Medium 1 and medium 2 are dielectrics with e., \varepsilon_{1}<\varepsilon_{1}.
(c) Medium 2 is a perfect conductor
(d) Impossible
(e) Medium 1 is a perfect conductor.
Fig.1 D N_{1}-D N_{2}=P_{S}
\begin{aligned}&\text { So } D N_{1}=P S . \Rightarrow E_{N 1}=P_{S} / E \\&E_{t 1}=E_{t 2}=0 \\&E_{1}=E_{t 1}+E_{N 1}=0+P_{S} / E=P_{S} / E\end{aligned}
Fig.2 D t_{2}>D t_{1}
\begin{aligned}&\text { As, } E_{t 1}=E_{t 2} \\&E_{2}>E_{1}\end{aligned}
Fig.3 \varepsilon_{1}>E_{1}
Fig.4 H_{1}=0, J=H_{2} A / m
Fig.5 E_{N 2}>E_{N 1}, E t_{1} \uparrow E t_{2}.
which is not a possible, boundary condition.
V_{3}=\frac{Q}{4 \pi E_{x}}+\frac{2 Q}{4 \pi E(x-50)} (3)
at equilibrium \text { P.E. }=Q V_{1}+2 Q V_{2}+Q_{3} V_{3}=0 from equation (1) (2) and (3)
\begin{aligned}&A\left[\frac{2 Q}{4 \pi E(50)}+\frac{Q_{3}}{4 \pi E(x)}\right] \\&+2 Q\left[\frac{Q}{4 \pi E(50)}+\frac{Q_{3}}{4 \pi E(x-50)}\right]\end{aligned}
+Q_{3}\left[\frac{Q}{4 \pi E(x)}+\frac{2 Q}{4 \pi E(x-50)}\right]=0 (4)
\text { Solving equation (4), } Q=-50 Q^{3}or Q_{3}=\frac{-1}{50} Q
so, x(x-50)=(x+25) \Rightarrow x^{2}-50 x-x+25=0 (5)
\text { Solving (5) } x=\frac{-51 \pm \sqrt{(-51)^{2}-4(1)(25)}}{2}= 30.5 of 0.5 cm
x = 0.5 cm as (0.5 << 50.5)
So, x = 50.5 cm.