A parallel-plate air capacitor has area A = 26.0 cm² and plate separation d = 3.20 mm, and it is charged to V = 54.0 V. If a dielectric with dielectric constant κ = 4.60 is inserted to completely fill the space between the plates while the capacitor remains connected to the battery, how much additional charge flows from the battery onto the positive plate?

What changes when the dielectric is inserted (battery stays connected)

Because the capacitor remains connected to the battery, the potential difference is fixed at $V=54.0\,\text{V}$. Inserting a dielectric increases the capacitance, so the stored charge must increase, and the extra charge comes from the battery.

Compute the original air capacitance

For a parallel-plate capacitor, $$C_0=\varepsilon_0\frac{A}{d}$$ Convert units:

• $A=26.0\,\text{cm}^2=26.0\times10^{-4}\,\text{m}^2=2.60\times10^{-3}\,\text{m}^2$
• $d=3.20\,\text{mm}=3.20\times10^{-3}\,\text{m}$

Now calculate: $$C_0=(8.854\times10^{-12})\frac{2.60\times10^{-3}}{3.20\times10^{-3}}\approx7.20\times10^{-12}\,\text{F}$$

Use $\Delta Q=(\kappa-1)C_0V$

With dielectric constant $\kappa$, the new capacitance is $C=\kappa C_0$, so $$\Delta Q=Q_f-Q_0=(\kappa C_0V)-(C_0V)=(\kappa-1)C_0V$$ Substitute values: $$\Delta Q=(4.60-1)(7.20\times10^{-12})(54.0)\approx1.40\times10^{-9}\,\text{C}$$

So about $1.40\,\text{nC}$ of additional charge flows onto the positive plate.

Quick reasonableness check

$\kappa=4.60$ means the charge should increase by a factor of $4.60$ at the same $V$, so the increase should be $3.60$ times the original charge, which matches $(\kappa-1)$ in the formula.