For capacitors in series, the equivalent capacitance is:

$\overline{)\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{e}\mathbf{q}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{n}}}}$

or

For two capacitors:

$\overline{){{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}\frac{{\mathbf{C}}_{\mathbf{1}}{\mathbf{C}}_{\mathbf{2}}}{{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}}}$

The capacitance of a parallel plate capacitor:

$\overline{){\mathit{C}}{\mathbf{=}}\frac{\mathit{K}{\mathit{\epsilon}}_{\mathbf{0}}\mathit{A}}{\mathit{d}}}$

Two parallel plate capacitors are connected in series. Both are filled with a dielectric with constant K.One capacitor had plates of area A_{1} separated by distance d_{1}, the other has plate of area A_{2} separated by distance d_{2}. What is the total capacitance of the two connected in series?

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