Question

In the case that the electric field E₀ is confined to the space between the plates of an isolated and charged parallel-plate capacitor, an inserted linear dielectric that fills the space would reduce E₀ and as a consequence the potential difference ΔVv

between them, by a factor of 1/κ, where κ
is the dielectric constant, because |E|=E
and ΔV are related linearly as ΔV=E×d. The charge on either plate is unaffected, simply because it has nowhere to go.

Hence, because ΔV₀→1/κ×ΔV₀, we have C₀=Q₀/ΔV₀→κ×C0

Now if the voltage between the plates is held fixed with the use of a source like a battery, regardless of the dielectric constant, ΔV₀ (and hence E₀=ΔV₀/d) are unchaged. The capacitance still increases by a factor of κ, and the explanation is that in this case, Q₀→κ×Q₀, with the additional charge coming from the wires attached to either plate.
What is the physical explanation behind this? Why does the charge have to increase on either plate?

Answer 1

The presence of a dielectric in a capacitor reduces the net electric field, leading to more charge accumulation on the plates to maintain a constant voltage when the capacitor is connected to a battery. This increases the total charge and capacitance of the capacitor by a factor of the dielectric constant.

Step-by-step explanation:

Dielectric Insertion in a Charged Capacitor

When a linear dielectric is inserted between the plates of an isolated charged parallel-plate capacitor that is not connected to a battery, the dielectric material reduces the electric field inside the capacitor because it becomes polarized. The alignment of charges within the dielectric opposes the original electric field, leading to a reduced net field. Consequently, the potential difference ΔV across the plates drops, which increases the capacitance (C = Q/ΔV). If the capacitor is connected to a battery that maintains a fixed voltage ΔV₀, the electric field E₀ remains constant. However, the presence of the dielectric allows the capacitor to store more charge (Q₀ → κ × Q₀) as the dielectric’s polarization reduces the effective field within the capacitor, permitting the addition of more free charge from the battery to achieve the same electric field strength.

The physical reason for the increase in charge on the plates is related to the dielectric's influence on the electric field. The dielectric constant, κ, quantifies how much an electric field is reduced inside a material compared to a vacuum. When the voltage is held constant by a battery, the reduced field inside the dielectric means that more charge must accumulate on the plates to maintain the original field strength. This extra charge flows in from the external circuit attached to the battery, increasing the total charge on the plates, and hence the capacitance of the capacitor, by a factor of κ.