Question

If I place a random arrangement of dielectrics like this in my capacitor, how do I find my new capacitance?

• A. $C=\frac{k_1{k}_2{k}_3{k}_4{\epsilon }_0{A}_2{A}_4}{k_2{d}_1+k_1{d}_2+k_4{d}_1+k_3{d}_2}$
• B. $C=\frac{k_1{k}_2{\epsilon }_0{A}_2}{k_2{d}_1+k_1{d}_2}+\frac{k_1{k}_4{\epsilon }_0(A_1-A_2)}{k_4{d}_1+k_1{d}_2}+\frac{k_4{k}_3{\epsilon }_0{A}_3}{k_4{d}_1+k_3{d}_2}$
• C. $C=\frac{{\epsilon }_0(k_1{k}_2+k_3{k}_4)(A_2+A_4+A_1+A_3)}{k_2{d}_1+k_1{d}_2+k_4{d}_1+k_3{d}_2}$
• D. None of these

Answer 1

The correct option is B $C=\frac{k_1{k}_2{\epsilon }_0{A}_2}{k_2{d}_1+k_1{d}_2}+\frac{k_1{k}_4{\epsilon }_0(A_1-A_2)}{k_4{d}_1+k_1{d}_2}+\frac{k_4{k}_3{\epsilon }_0{A}_3}{k_4{d}_1+k_3{d}_2}$

It will be easier and you will never go wrong if you split this into a set of parallel dielectric combinations, each having 2 series dielectrics.

So we have split it into 3 parallel capacitors as shown,
$C=C_1+C_2+C_3$
Now each of these is a series combination,
$c_1=\frac{1}{\frac{1}{\frac{k_1{\epsilon }_0{A}_2}{d_1}}+\frac{1}{\frac{k_2{\epsilon }_0{A}_2}{d_2}}}$
Similarly, if you calculate $C_2$ and $C_3$ and sum them up, you get
$C=\frac{k_1{k}_2{\epsilon }_0{A}_2}{k_2{d}_1+k_1{d}_2}+\frac{k_1{k}_4{\epsilon }_0(A_1-A_2)}{k_4{d}_1+k_1{d}_2}+\frac{k_4{k}_3{\epsilon }_0{A}_3}{k_4{d}_1+k_3{d}_2}$