Question

Find the equivalent capacitance between P and Q in the following diagram. The area of each plate is A and the separation between plates is d.

Answer 1

(i) The capacitor in the first diagram is a series combination of three capacitors of plate separations d/3 and plate areas A, with C₁ filled with air (k₁= 1), C₂ filled with a dielectric of k₂ = 3, and C₃ filled with a dielectric of k₃ = 6.

∴ ${\displaystyle {\text{C}}_1=\frac{{\text{k}}_1{\epsilon }_0\text{A}}{\text{d}/3}=\frac{3{\epsilon }_0\text{A}}{\text{d}}{\text{k}}_1,{\text{C}}_2=\frac{{\text{k}}_2{\epsilon }_0\text{A}}{\text{d}/3}=\frac{3{\epsilon }_0\text{A}}{\text{d}}{\text{k}}_2}$,

${\displaystyle {\text{C}}_3=\frac{{\text{k}}_3{\epsilon }_0\text{A}}{\text{d}/3}=\frac{3{\epsilon }_0\text{A}}{\text{d}}{\text{k}}_3}$

∴ ${\displaystyle \frac{1}{\text{C'}}=\frac{1}{{\text{C}}_1}+\frac{1}{{\text{C}}_2}+\frac{1}{{\text{C}}_3}=\frac{\text{d}}{3{\epsilon }_0\text{A}}(\frac{1}{{\text{k}}_1}+\frac{1}{{\text{k}}_2}+\frac{1}{{\text{k}}_3})}$

∴ C' = ${\displaystyle \frac{3{\epsilon }_0\text{A}}{\text{d}}\left(\frac{{\text{k}}_1{\text{k}}_2{\text{k}}_3}{{\text{k}}_1{\text{k}}_2+{\text{k}}_2{\text{k}}_3+{\text{k}}_3{\text{k}}_1}\right)=\frac{3{\epsilon }_0\text{A}}{\text{d}}\left(\frac{18}{27}\right)}$

= ${\displaystyle \frac{2{\epsilon }_0\text{A}}{\text{d}}}$

(ii) A series combination of two capacitors  C₂ (k₂ = 3) and C₃ (k₃ = 6) of plate areas A/2 and plate separations d/2 is in parallel with a capactor C₁ (k₁ = 4) of plate area A/2 and plate separation d in the second figure above.

∴ ${\displaystyle {\text{C}}_1=\frac{{\text{k}}_1{\epsilon }_0(\text{A}/2)}{\text{d}}=\frac{{\epsilon }_0\text{A}}{\text{2d}}{\text{k}}_1}$

${\displaystyle {\text{C}}_2=\frac{{\text{k}}_2{\epsilon }_0(\text{A}/2)}{\text{d}/2}=\frac{{\epsilon }_0\text{A}}{\text{d}}{\text{k}}_2}$

${\displaystyle {\text{C}}_3=\frac{{\text{k}}_3{\epsilon }_0(\text{A}/2)}{\text{d}/2}=\frac{{\epsilon }_0\text{A}}{\text{d}}{\text{k}}_3}$

∴ For the series combination of C₂ and C₃,

${\displaystyle \frac{1}{{\text{C}}_4}=\frac{1}{{\text{C}}_2}+\frac{1}{{\text{C}}_3}}$

∴ ${\displaystyle {\text{C}}_4=\frac{{\text{C}}_2{\text{C}}_3}{{\text{C}}_2+{\text{C}}_3}=\frac{{\epsilon }_0\text{A}}{\text{d}}\left(\frac{{\text{k}}_2{\text{k}}_3}{{\text{k}}_2+{\text{k}}_3}\right)=\frac{{\epsilon }_0\text{A}}{\text{d}}\left(\frac{3\times 6}{3+6}\right)=\frac{2{\epsilon }_0\text{A}}{\text{d}}}$

Finally, for the parallel combination of C₁ and C₄, 

C'' = ${\displaystyle {\text{C}}_1+{\text{C}}_4=\frac{{\epsilon }_0\text{A}}{\text{2d}}\left(4\right)+\frac{2{\epsilon }_0\text{A}}{\text{d}}=\frac{4{\epsilon }_0\text{A}}{\text{d}}}$

Thus, the equivalent capacitances are C' = ${\displaystyle \frac{2{\epsilon }_0\text{A}}{\text{d}}}$ and C'' = ${\displaystyle \frac{4{\epsilon }_0\text{A}}{\text{d}}}$.