Electrostatics 5 Question 15
17. A charged particle is free to move in an electric field. Will it always move along an electric line of force?
(1979)
(d) $\frac{E _1}{E _2}=\frac{\left(K _1+K _2+K _3\right)\left(K _2 K _3+K _3 K _1+K _1 K _2\right)}{9 K _1 K _2 K _3}$
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Solution:
- Let’s consider a strip of thickness ’ $d x$ ’ at a distance of ’ $x$ ’ from the left end as shown in the figure. From the figure, $\triangle A B C$ and $\triangle A D E$ are similar triangles,
$$ \Rightarrow \quad \frac{y}{x}=\frac{d}{a} \Rightarrow y=\frac{d}{a} x $$
We know that, the capacitance of parallel plate capacitor,
$$ \begin{aligned} C & =\frac{\varepsilon _0 A}{d} \\ C _1=\frac{\varepsilon _0(a d x)}{(d-y)} \text { and } C _2 & =\frac{K \varepsilon _0(a d x)}{y} \end{aligned} $$
Here, two capacitor are placed in series with variable thickness, therefore
$$ C _{eq}=\frac{C _1 C _2}{C _1+C _2} \Rightarrow C _{eq}=\frac{K \varepsilon _0 a d x}{K d+(1-K) y} $$
Now, integrate it from 0 to $a$
$$ C=\int _0^{a} \frac{K \varepsilon _0 a d x}{K d+(1-K) y} $$
Using Eq. (i), $y=\frac{d}{a} x$, we get
$$ \begin{aligned} \quad C & =\varepsilon _0 a \int _0^{a} \frac{d x}{d+\frac{1}{K}-1 \frac{d}{a} x} \\ \Rightarrow \quad C & =\frac{\varepsilon _0 a}{\frac{1-K}{K} \frac{d}{a}} \ln \frac{1}{K} \\ \Rightarrow \quad C & =\frac{\varepsilon _0 a^{2} K \ln K}{(K-1) d} \end{aligned} $$
