A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that co

Q: Why are the two regions considered to be in parallel?

Because the top and bottom regions share the same continuous metal plates on the left and right, the potential difference across both regions is forced to be identical. This is the defining characteristic of a parallel combination.

Q: Doesn't the dielectric reduce the electric field? Why is $E_1 = E_2$?

In an isolated capacitor with a fixed charge, inserting a dielectric reduces the electric field. However, here the capacitor is connected to a voltage source (or the plates are equipotential surfaces), so the potential difference $V$ is fixed. Since $E = V/d$ and both $V$ and $d$ are the same for both regions, the electric field must be identical.

Q: How do we determine the area of each region?

The problem states the dielectric covers $1/3$ of the plate area. Therefore, the area of the dielectric region is $A/3$, leaving the remaining $2/3$ of the area ($2A/3$) for the air-filled region.

Q: Why is the charge not distributed equally?

Charge distribution depends on capacitance ($Q = CV$). Since the dielectric region has a higher capacitance per unit area (due to the dielectric constant $K$), it will draw proportionally more charge than the air-filled region.

Q: What does 'ignoring edge effects' mean?

In reality, the electric field lines bulge outward at the edges of the plates and at the boundary between the dielectric and air. Ignoring these 'fringing fields' allows us to assume the electric field is perfectly uniform and perpendicular to the plates everywhere.

Question

A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $1/3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_1$. When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. The electric field in the dielectric is $E_1$ and that in the other portion is $E_2$. Choose the correct option/options, ignoring edge effects.

Options:

(A) $\frac{E_1}{E_2} = 1$

(B) $\frac{E_1}{E_2} = \frac{1}{K}$

(C) $\frac{Q_1}{Q_2} = \frac{3}{K}$

(D) $\frac{C}{C_1} = \frac{2+K}{K}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a standard parallel plate capacitor, but with a twist. A dielectric slab of dielectric constant $K$ has been inserted, but it doesn't fill the entire space. Instead, it only covers the upper $1/3$ of the area of the plates.

Because the left and right metal plates are continuous and highly conductive, they act as equipotential surfaces. This means the potential difference $V$ across the upper region (with the dielectric) is exactly the same as the potential difference across the lower region (with air). Whenever two components share the same potential difference, they are in parallel. Therefore, we can elegantly model this system as two separate capacitors connected in parallel!

### The Electric Field Mystery

One of the most common traps in electrostatics is assuming that a dielectric always reduces the electric field. Let's clear that up. The electric field $E$ between two parallel plates is fundamentally given by the potential difference divided by the separation distance:

$$E = \frac{V}{d}$$

In our setup, both the upper and lower regions share the exact same potential difference $V$ and the exact same plate separation $d$. Because neither $V$ nor $d$ changes between the two regions, the electric field must be perfectly uniform throughout the entire space between the plates.

Therefore, the electric field in the dielectric ($E_1$) is equal to the electric field in the air gap ($E_2$).

$$E_1 = E_2 \implies \frac{E_1}{E_2} = 1$$

This immediately tells us that option (a) is correct and option (b) is incorrect.

### Calculating the Capacitances

Now, let's determine the individual capacitances of our two imaginary parallel capacitors. The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon A}{d}$.

For the upper region ($C_1$), the area is $A/3$ and the permittivity is $K\varepsilon_0$:

$$C_1 = \frac{K \varepsilon_0 (A/3)}{d} = \frac{K \varepsilon_0 A}{3d}$$

For the lower region ($C_2$), the area is the remaining $2A/3$ and the permittivity is just $\varepsilon_0$ (since it's air):

$$C_2 = \frac{\varepsilon_0 (2A/3)}{d} = \frac{2 \varepsilon_0 A}{3d}$$

### The Master Equation for Total Capacitance

Since $C_1$ and $C_2$ are in parallel, their equivalent total capacitance $C$ is simply their sum:

$$C = C_1 + C_2$$

Substituting our expressions:

$$C = \frac{K \varepsilon_0 A}{3d} + \frac{2 \varepsilon_0 A}{3d} = \frac{(K+2) \varepsilon_0 A}{3d}$$

To check option (d), we need the ratio of the total capacitance $C$ to the upper capacitance $C_1$:

$$\frac{C}{C_1} = \frac{\frac{(K+2) \varepsilon_0 A}{3d}}{\frac{K \varepsilon_0 A}{3d}}$$

The common term $\frac{\varepsilon_0 A}{3d}$ beautifully cancels out, leaving us with:

$$\frac{C}{C_1} = \frac{K+2}{K}$$

This perfectly matches option (d)!

### Final Verification: The Charge Ratio

Just to be absolutely thorough, let's check the charge ratio proposed in option (c). The charge stored on a capacitor is the product of its capacitance and the voltage across it ($Q = CV$).

Since the voltage $V$ is identical for both regions, the ratio of their charges is simply the ratio of their capacitances:

$$\frac{Q_1}{Q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2}$$

Substituting our capacitance values:

$$\frac{Q_1}{Q_2} = \frac{\frac{K \varepsilon_0 A}{3d}}{\frac{2 \varepsilon_0 A}{3d}} = \frac{K}{2}$$

Option (c) claims the ratio is $3/K$, which is clearly incorrect.

**Conclusion:** The correct options are indeed **(a)** and **(d)**. This problem is a fantastic exercise in recognizing parallel configurations and understanding the true dependencies of the electric field!

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Visualized Solution

$Parallel Plate Model$

$Electric Field in Parallel Regions$

$Individual Capacitances$

$Equivalent Capacitance$

$Ratio of Capacitances$

$Ratio of Charges$

$Conclusion$

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Electric Field Mystery

Calculating the Capacitances

The Master Equation for Total Capacitance

Final Verification

Similar Questions

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A parallel plate capacitor with plates of area $1 m^{2}$ each, are at a separation of $0.1 m$ . If the electric field between the plates is $100 N/C$ , the magnitude of charge on each plate is $(Take, ε_{0} = 8.85 \times 1 0^{- 12} \frac{C ^{2}}{N-m ^{2}})$

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Plate Model

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field in Parallel Regions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Individual Capacitances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Capacitance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Capacitances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Charges

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion