Question

A parallel plate capacitor is of area 3A
and plates separation 3d. It is filled with
three dielectric slabs with dielectric
W
constant K1 = 2; K2 = 4 ad K3 = 6 as
shown in figure. The capacitance of the capacitor is:
17
25
2A
Q
Кі
K
А
20
A
A
B​

Answer 1

Answer:

The capacitance of the capacitor is $\frac{2}{\mathrm{k}}=\frac{3}{\mathrm{k}_1+\mathrm{k}_2+\mathrm{k}_5}+\frac{1}{\mathrm{k}_4}$

Explanation:

Step 1: Write the capacitance of all parts of capacitor filled with dielectries

Let $\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3$and $\mathrm{C}_4$ are the capacitance of parts of the capacitor filled with dielectric constant $\mathrm{k}_1, \mathrm{k}_2, \mathrm{k}_3 and \mathrm{k}_4$

$\mathrm{C}_1=\frac{(\mathrm{A} / 3) \mathrm{k}_1 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_1}{3 \mathrm{k}} \mathrm{C},$

$\mathrm{C}_2=\frac{(\mathrm{A} / 3) \mathrm{k}_2 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2}{3 \mathrm{k}} \mathrm{C},$

$\mathrm{C}_3=\frac{(\mathrm{A} / 3) \mathrm{k}_3 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_1}{3 \mathrm{k}} \mathrm{C},$

$\mathrm{C}_4=\frac{(\mathrm{A}) \mathrm{k}_4 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_4}{\mathrm{k}} \mathrm{C}$

Step 2: Calculate equivalent capacitance of arrangement

The above three capacitors $\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3$ are in parallel and then it is in series with $\mathrm{C}_4$Now the equivalent capacitance for the combination of four capacitors is

$\begin{aligned}& 1 / \mathrm{C}_{\mathrm{eq}}=1 /\left(\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3\right)+1 / \mathrm{C}_4 \\& \text { or } 1 / \mathrm{C}_{\mathrm{eq}}=\frac{3 \mathrm{k}}{2 \mathrm{C}}\left[\frac{1}{\mathrm{k}_1+\mathrm{k}_2+\mathrm{k}_3}\right]+\frac{\mathrm{k}}{2 \mathrm{k}_4 \mathrm{C}}\end{aligned}$

Step 3:Equating this equivalent capacitance to the capacitance of capacitor filled with single dielectric

$\begin{aligned}\mathrm{C} & =\mathrm{C}_{\mathrm{e}} \mathrm{q} \\\frac{1}{\mathrm{C}} & \left.=\frac{1}{\mathrm{C}_{\mathrm{e}} \mathrm{q}} \quad \text { (Put } \mathrm{C}=\frac{\mathrm{k} \epsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}\right)\end{aligned}$

$\text { or } \frac{2}{\mathrm{k}}=\frac{3}{\mathrm{k}_1+\mathrm{k}_2+\mathrm{k}_3}+\frac{1}{\mathrm{k}_4}$

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