Question 567: Electrostatics & Capacitance

Question

Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [A = area of plate. $$t_1, t_2$$ and $$t_3$$ are thickness of dielectric slabs, $$\mathrm{k}_1, \mathrm{k}_2$$ and $$\mathrm{k}_3$$ are dielectric constants.

Accepted Answer

In the absence of the dielectric the electric field between the plates is given by
$$\mathrm{E}=\frac{\mathrm{q}}{\mathrm{A} \varepsilon_0}$$
when dielectric is placed, the field gets reduced by the factor $$k$$.
The potential difference between the plates is given by
$$\begin{aligned}
V & =E_1 t_1+E_2 t_2+E_2 t_3 \
& =\frac{E}{k_1} \cdot t_1+\frac{E}{k_2} \cdot t_2+\frac{E}{k_3} \cdot t_3 \
& =E\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}ight) \
& =\frac{q}{A \varepsilon_0}\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}ight) \
C & =\frac{q}{V}=\frac{A \varepsilon_0}{\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}ight)}
\end{aligned}$$

Answer

$$\frac{A \varepsilon_0}{\left[\frac{t_1+t_2+t_3}{k_1+k_2+k_3}\right]}$$

Answer

$$
\frac{A \varepsilon_0\left(k_1 k_2 k_3\right)}{t_1 t_2 t_3}
$$

Answer

$$
A \varepsilon_0\left[\frac{k_1}{t_1}+\frac{k_2}{t_2}+\frac{k_3}{t_3}\right]
$$

Answer

$$
\frac{A \varepsilon_0}{\left[\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}\right]}
$$

Electrostatics & Capacitance

Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [A = area of plate. $t_1, t_2 and t_3 are thickness of dielectric slabs, \mathrm{k}_1, \mathrm{k}_2 and \mathrm{k}_3$ are dielectric constants.

Choose the correct answer: