electrostatic-potential-and-capacitance Question 30
Question: Q. 5. Two identical capacitors of plate dimension $l \times b$ and plate separation $d$ have dielectric slabs filled in between the space of the plates as shown in the figures.
Obtain the relation between the dielectric constants $\kappa, \kappa_{1}$ and $\kappa_{2}$.
A [O.D. Comptt. I, II, III 2013]
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Solution:
Ans. The capacitor can be considered as split into two capacitors connected in parallel.
Here $\quad C_{1}=\frac{\kappa_{1} \varepsilon_{0} \mathrm{~A} / 2}{d}, C_{2}=\frac{\kappa_{2} \varepsilon_{0} \mathrm{~A} / 2}{d}$
$1 / 2$
In parallel combination,
$$ \begin{align*} & C_{e q}=C_{1}+C_{2} \ & C_{e q}=\frac{\kappa_{1} \varepsilon_{0} \mathrm{~A}}{2 d}+\frac{\kappa_{2} \varepsilon_{0} \mathrm{~A}}{2 d} \end{align*} $$
From this figure
$$ \begin{equation*} \mathrm{C}{e q}=\frac{\kappa \varepsilon{0} \mathrm{~A}}{2 d} \tag{ii} \end{equation*} $$
From (i) and (ii)
$$ \begin{aligned} \frac{\kappa \varepsilon_{0} \mathrm{~A}}{2 d} & =\frac{\varepsilon_{0} \mathrm{~A}}{2 d}\left(\kappa_{1}+\kappa_{2}\right) \ \kappa & =\kappa_{1}+\kappa_{2} \end{aligned} $$
