The Initial State
A Fully Filled Capacitor
Imagine a standard parallel plate capacitor. Initially, the entire space between its plates, a distance d apart, is completely filled with a dielectric material of constant K. The capacitor is hooked up to a battery that maintains a constant voltage V across the plates.
In this initial configuration, the capacitance is straightforward. It is given by the standard formula for a filled capacitor:
Because the battery is connected, the initial electric field E0 inside the dielectric is simply the voltage divided by the distance:
The Transformation
Pulling the Plates Apart
Now, let's perform a thought experiment. We grab both plates and pull them outward, each by a distance of 2d. The dielectric slab, however, stays right where it is in the middle.
What does our new system look like? We now have three distinct regions between the plates: an air gap of width 2d, the original dielectric slab of width d, and another air gap of width 2d.
Because the electric field lines must pass sequentially through these three layers, this physical arrangement is electrically equivalent to three capacitors connected in series.
Calculating the New Capacitance
To find the new equivalent capacitance C′, we use the general formula for a capacitor with multiple dielectric layers:
Let's substitute the thickness (ti) and dielectric constant (Ki) for each of our three layers:
Notice that the two air gaps add up perfectly: 2d+2d=d. This simplifies our denominator beautifully:
C′=d+Kdε0A=d(1+K1)ε0A
Taking the common denominator in the bracket yields KK+1. Flipping the K to the numerator gives us our final expression:
Do you recognize the term dKε0A? That is exactly our initial capacitance C! Therefore, we can write:
This tells us that the new capacitance is the original capacitance multiplied by K+11. In the phrasing of the options, the capacitance is decreased by a factor of K+11. This makes Option (B) the correct choice.
The Electric Field Mystery
What happens to the electric field inside the dielectric? Since the battery remains connected throughout the process, the total potential difference V across the plates is locked and constant.
This total voltage is the sum of the potential drops across the air gaps and the dielectric:
V=Eair⋅dair+Edielectric⋅ddielectric
We know the total air gap is d, and the dielectric thickness is also d. But how do Eair and Edielectric relate? From Gauss's Law, the electric displacement field D is continuous across the boundary of the dielectric. This means:
Substituting this powerful relation back into our voltage equation:
V=(KEdielectric)⋅d+Edielectric⋅d
Factoring out Edielectric⋅d, we get:
Solving for the new electric field inside the dielectric:
Since our initial electric field was E0=dV, we can clearly see that:
The electric field is reduced by a factor of (K+1), not 2K. Thus, Option (A) is incorrect.
The Energy and Work Done
Finally, let's address the work done. When we pull the plates apart, the capacitance changes while the voltage remains constant. This means the battery must do work to move charge, and the stored potential energy of the capacitor changes.
The work done by the external agent is the difference between the change in stored energy and the work done by the battery:
Wext=ΔU−Wbattery=21(C′−C)V2−(C′−C)V2=−21(C′−C)V2
Because the new capacitance C′ explicitly depends on the dielectric constant K, the term (C′−C) depends on K. Consequently, the work done absolutely depends on the presence of the dielectric material. Option (D) is therefore incorrect.
