Question

A vacuum-insulated parallel-plate capacitor with plate separation $d$has capacitance $C_0$. What is the capacitance if an insulator with dielectric constant $\kappa$and thickness $d/2$is slipped between the electrodes without changing the plate separation?

Short answer

If an insulator with dielectric constant $\kappa$and thickness $d/2$is slipped between the electrodes without changing the plate separation then the capacitance is $C_s=\frac{2\kappa C_0}{\kappa +1}$

Step-by-step solution

Given Information

Parallel-plate capacitor with plate separation $=d$

Capacitance$=C_0$

Dielectric constant$=k$

Thickness$=d/2$

Explanation

If we insert an insulator into such a capacitor with separation $d$and $k=1$, then we can treat it as two capacitors connected in series, each with thickness $d/2$and $\kappa =1$

It is clear that the capacitance of the air half will be

$C_1=2C_0$

Since the capacitance is given as

$C={\epsilon }_0\kappa \frac{A}{d}.$

The capacitance of the other half will then clearly be

$C_2=2\kappa C_0.$

Resulting capacitance of the considered system will be

$C_s={\left(\frac{1}{2C_0}+\frac{1}{2\kappa C_0}\right)}^{-1}={\left(\frac{\kappa +1}{2\kappa C_0}\right)}^{-1}$

The resulting capacitance will be

role="math" localid="1648714430037" $C_s=\left(\frac{2\kappa C_0}{\kappa +1}\right)$

Final Answer

Hence, the capacitance is $C_s=\frac{2\kappa C_0}{\kappa +1}.$