Effect on a capacitor due to a dielectric

Question

Why does the effect of a dielectric fully inserted between a parallel plate capacitor differ when the capacitor is connected to a battery source and when it is not?

When a charged capacitor is not connected to a battery, its potential difference can be defined as $$V=\frac{V_\text{initial}}{K}$$ where $V_\text{initial}$ refers to the potential difference prior to the insertion of the dielectric and $K$ is the dielectric constant.

However, when the capacitor is connected to a battery source, the potential difference remains the same, even when a dielectric is inserted inside the capacitors. What is the reason for this?

Answer 1

When the capacitor is not connected to anything, the free charge on its plates have nowhere to move to so it remains the same. When the dielectric is inserted between the plates, the positive plate attracts the negative charges towards itself and vice versa. This causes a net displacement of charge and thus polarizes the dielectric in the same direction as the applied electric field. This produces accumulations of bound charge at either end that are of opposite sign as the free charge. This decreases the total charge density at the plates and therefore the potential difference.

On the other hand, an ideal battery by definition maintains a certain potential difference between its terminals. Therefore, when the dielectric is inserted, it will do additional work by increasing the charge until the potential difference is reached.

Answer 2

However, when the capacitor is connected to a battery source, the potential difference remains the same, even when a dielectric is inserted inside the capacitors. What is the reason for this?

The voltage (potential difference) across the capacitor does initially change. It temporarily drops. Given the relationship

$$C=\frac{Q}{V}$$

Insertion of the dielectric increases the capacitance. Since the charge cannot change instantaneously, the increase in capacitance must be accompanied by a decrease in the capacitor voltage to below the battery emf (open circuit voltage). This causes the battery to charge the capacitor until its voltage equals the battery emf.

Keep in mind that all real batteries have internal resistance. So the battery charging current results in a voltage drop across the battery internal resistance. This means that during charging the battery terminal voltage is below the battery emf and matches the capacitor voltage. When the charging is complete the battery terminal voltage and capacitor voltage will equal the battery emf.

Hope this helps.