Animated Solution for Physics - Electrostatics: If $q_f$ is the free charge on the capacitor plates and $q_b$ is the bound charge on the dielectric slab of dielectric constant $K$ placed between the capacitor plates, then bound charge $q_b$ can be expressed as

Free Charges and the External Field

Imagine a parallel plate capacitor connected to a battery. The battery pumps electrons, creating a surplus of negative charge on one plate and a deficit on the other. These charges, which reside on the metal plates and can move freely if given a conducting path, are called free charges ($q_f$).

Because we have a separation of positive and negative charges, an electric field is established in the space between the plates. We call this the external electric field ($E_0$). From Gauss's Law, we know that the magnitude of this field in a vacuum is directly proportional to the surface charge density of the free charges:

Polarization and Bound Charges

Now, let's introduce a dielectric slab into this space. A dielectric is an insulator; its electrons are tightly bound to their parent atoms. However, when subjected to the external field $E_0$, these atoms experience a stretching force. The positive nuclei are pulled slightly in the direction of the field, while the negative electron clouds are pushed in the opposite direction.

This microscopic stretching is called polarization. Macroscopically, the internal charges cancel each other out, but at the surfaces of the dielectric, a net charge appears. These are the bound charges ($q_b$). They are "bound" because they cannot flow away; they are merely the exposed ends of the polarized atoms.

Crucially, these bound charges create their own electric field, the induced electric field ($E_p$), which points from the positive bound charge to the negative bound charge. Notice that $E_p$ points in the exact opposite direction to the external field $E_0$!

Net Electric Field

Inside the dielectric, we now have a tug-of-war. The external field $E_0$ is pushing one way, and the induced field $E_p$ is pushing the other. The net electric field ($E_{net}$) is the vector sum of the two. Since they are anti-parallel, we simply subtract their magnitudes:

Substituting our expressions for the fields, we get:

We also have another way to define the net electric field. The dielectric constant ($K$) of a material is fundamentally defined as the factor by which the material reduces the external electric field. Therefore, the net field can also be written as:

Now we have two different expressions for the exact same physical quantity ($E_{net}$). Let's equate them to unlock the relationship between the free and bound charges:

This equation looks a bit cluttered, but notice that every single term is divided by the area $A$ and the permittivity of free space ${\epsilon }_0$. We can multiply the entire equation by $A{\epsilon }_0$ to cancel these terms out, leaving us with a beautifully simple algebraic relation:

All that's left is to isolate our target variable, the bound charge $q_b$. By rearranging the terms, we get:

Factoring out the free charge $q_f$, we arrive at our final, elegant formula:

This perfectly matches option (b). But let's not stop at the math; let's push this formula to its limits to see if it makes physical sense.

What if the slab wasn't a dielectric, but a perfect metal conductor? Inside a perfect conductor in electrostatics, the net electric field must be exactly zero. For $E_{net}$ to be zero, the induced field must perfectly cancel the external field ($E_p=E_0$), which means the bound charge must equal the free charge ($q_b=q_f$).

Does our formula predict this? For a perfect conductor, the dielectric constant $K\to \infty$. If we plug $K=\infty$ into our formula, the term $1/K$ becomes zero, and we are left with $q_b=q_f(1-0)=q_f$. The math perfectly aligns with the physical reality!