Equivalent capacitance for capacitors:

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

The capacitance of a parallel plate capacitor:

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

For a vacuum, k = 1. For other materials, it is greater than 1.

Before inserting a dielectric, the capacitance of the capacitor is

A vacuum-insulated parallel-plate capacitor with plate separation *d* has capacitance *C*_{0}.

What is the capacitance if an insulator with dielectric constant *κ* and thickness *d/*2 is slipped between the electrodes? Assume plate separation is unchanged.

Express your answer in terms of the variables *d*, *C*_{0}, and *κ*.

C = ?

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