electrostatic-potential-and-capacitance Question 29
Question: Q. 4. A slab of material of dielectric constant $\kappa$ has the same area as that of the plates of a parallel plate capacitor but has the thickness $d / 2$, where $d$ is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
U [O.D. I, II, III 2013]
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Solution:
Ans. Initially when there is vacuum between the two plates, the capacitance of the two parallel plates is,
$$ C_{0}=\frac{\varepsilon_{0} A}{d} $$
where, $A$ is the area of parallel plates.
Suppose that the capacitor is connected to a battery, an electric field $E_{0}$ is produced.
Now if we insert the dielectric slab of thickness $t=d / 2$, the electric field reduces to $E$.
Now the gap between the plates is divided in two parts, for distance $t$ there is electric field $E$ and for the remaining distance $(d-t)$, the electric field is $E_{0}$. $\quad 1 / 2$ If $V$ be the potential difference between the plates of the capacitor, then
$$ \begin{aligned} V & =E t+E_{0}(d-t) \ V & =\frac{E d}{2}+\frac{E_{0} d}{2} \ & =\frac{d}{2}\left(E+E_{0}\right) \ \Rightarrow \quad V & =\frac{d}{2}\left(\frac{E_{0}}{\kappa}+E_{0}\right) \ & =\frac{d E_{0}}{2 \kappa}(\kappa+1) \end{aligned} $$
$$ \begin{aligned} V & =\frac{E d}{2}+\frac{E_{0} d}{2} \quad\left(\because t=\frac{d}{2}\right) \ & =\frac{d}{2}\left(E+E_{0}\right) \quad\left(\because \frac{E_{0}}{E}=\kappa\right) \ \Rightarrow \quad V & =\frac{d}{2}\left(\frac{E_{0}}{\kappa}+E_{0}\right) \ & =\frac{d E_{0}}{2 \kappa}(\kappa+1) \ \text { Now, } \quad E_{0} & =\frac{\sigma}{\varepsilon_{0}}=\frac{q}{\varepsilon_{0} A} \ \Rightarrow \quad V & =\frac{d}{2 \kappa} \cdot \frac{q}{\varepsilon_{0} A}(\kappa+1) \end{aligned} $$
We know,
$$ C=\frac{q}{V}=\frac{2 \kappa \varepsilon_{0} A}{d(\kappa+1)} $$
