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Homework answers / question archive / A vacuum-insulated parallel-plate capacitor with plate separation d has capacitance C0

A vacuum-insulated parallel-plate capacitor with plate separation d has capacitance C0

Physics

A vacuum-insulated parallel-plate capacitor with plate separation d has capacitance C0.

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, C0, and κ.
C=?

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Capacitance is given by:

C_0=\frac{\varepsilon _0A}{d}

You can think of this as two capacitors in series if you assume that the material touches the bottom plate. The first capacitor has the dielectric material in it while the second has vacuum. Let's call these capacitors C1 and C2. Each one has a separation of d/2

C_1=\frac{\varepsilon _0kA}{d/2}=\frac{2\varepsilon _0kA}{d}=2kC_0

C_2=\frac{\varepsilon _0A}{d/2}=\frac{2\varepsilon _0A}{d}=2C_0

Since you are assuming they are in series, the total capacitance is found by:

\frac{1}{C_{total}}=\frac{1}{C_1}+\frac{1}{C_2}

Substituting C1 and C2

\frac{1}{C_{total}}=\frac{1}{2kC_0}+\frac{1}{2C_0}

solving for Ctotal:

C_{total}=\frac{4kC_0^2}{2C_0(1+k)}

C_{total}=\frac{2kC_0}{1+k}

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