A container has a base of 50 cm x 5 cm and height 50 cm, as shown in the figure. It has two parallel

Q: Why are the two parts considered to be in parallel and not in series?

The conducting walls (plates) are continuous from top to bottom. Both the liquid and the air are exposed to the same potential difference across these plates, which is the defining characteristic of a parallel combination.

Q: How did we identify which walls are the plates?

The problem states there are 'two parallel electrically conducting walls each of area $50 \text{ cm} \times 50 \text{ cm}$'. In a capacitor, the conducting surfaces act as the plates.

Q: Why did we use $50 \times 5$ for the base area?

The container's base is given as $50 \text{ cm} \times 5 \text{ cm}$. The liquid fills from the bottom up, so the volume it occupies is the base area multiplied by the height it reaches.

Q: Why is the area for the liquid capacitor $50 \times 10$?

The plates are the $50 \times 50$ walls. The liquid reaches a height of $10 \text{ cm}$. So, the portion of the plates in contact with the liquid has a width of $50 \text{ cm}$ and a height of $10 \text{ cm}$.

Q: What happens to the non-conducting walls?

The problem explicitly states 'the effects of the non-conducting walls on the capacitance are negligible'. They just serve to hold the liquid and don't participate in storing charge.

Question

A container has a base of $50 	ext{ cm} 	imes 5 	ext{ cm}$ and height $50 	ext{ cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 	ext{ cm} 	imes 50 	ext{ cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of $250 	ext{ cm}^3 	ext{ s}^{-1}$. What is the value of the capacitance of the container after 10 seconds?
[Given: Permittivity of free space $\epsilon_0 = 9 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]

Options:

(A) 27 pF

(B) 63 pF

(C) 81 pF

(D) 135 pF

Sigma Percentile · Accepted Answer

Imagine you are in a high-tech laboratory, and you have a rectangular container. But this isn't just any container; two of its opposite walls are made of conducting metal, turning the whole thing into a giant capacitor! Now, you start pouring a special dielectric liquid into it. As the liquid level rises, the capacitance of the system changes. This is a beautiful problem that blends basic fluid geometry with electrostatics.

### Analyzing the Setup

First, we need to clearly identify the geometry of our capacitor. The problem states that the container has two parallel electrically conducting walls, each with an area of $50 	ext{ cm} 	imes 50 	ext{ cm}$. These are our capacitor plates!

The base of the container is $50 	ext{ cm} 	imes 5 	ext{ cm}$. Since the plates are $50 	ext{ cm}$ wide, the distance between them must be the other dimension of the base, which is $d = 5 	ext{ cm}$.

### The Fluid Dynamics

We are pouring a liquid with a dielectric constant $K = 3$ into this container at a uniform rate of $250 	ext{ cm}^3 	ext{ s}^{-1}$. We need to find the situation exactly after $10 	ext{ seconds}$.

Let's find out how much liquid is in the tank.

$$V = 	ext{rate} 	imes 	ext{time}$$

$$V = 250 	imes 10 = 2500 	ext{ cm}^3$$

Now, how high does this liquid reach? The liquid fills the container from the bottom up. The base area is $50 	ext{ cm} 	imes 5 	ext{ cm} = 250 	ext{ cm}^2$.

$$h = \frac{	ext{Volume}}{	ext{Base Area}} = \frac{2500}{250} = 10 	ext{ cm}$$

So, the liquid forms a layer $10 	ext{ cm}$ high at the bottom of the $50 	ext{ cm}$ tall container.

### The Master Equation

Here is where the physics intuition kicks in. The liquid doesn't fill the entire space between the plates. We now have two distinct regions between our conducting walls: a bottom region filled with liquid ($10 	ext{ cm}$ high) and a top region filled with air ($40 	ext{ cm}$ high).

Because the conducting plates are continuous, both the liquid region and the air region are subjected to the exact same potential difference. This means they act as **two capacitors connected in parallel**.

The equivalent capacitance is simply the sum of the two:

$$C_{eq} = C_{	ext{liquid}} + C_{	ext{air}}$$

The formula for a parallel plate capacitor is $C = \frac{K \epsilon_0 A}{d}$.

### Final Calculation

Let's calculate the capacitance for the liquid part. The area of the plates in contact with the liquid is $50 	ext{ cm}$ (width) $	imes 10 	ext{ cm}$ (height). We must convert these to meters to avoid silly mistakes!

$$A_{	ext{liquid}} = 50 	imes 10 	imes 10^{-4} 	ext{ m}^2 = 500 	imes 10^{-4} 	ext{ m}^2$$

$$C_{	ext{liquid}} = \frac{3 \cdot \epsilon_0 \cdot (500 	imes 10^{-4})}{5 	imes 10^{-2}} = 3\epsilon_0$$

Now, for the air part ($K=1$). The height of the air column is $50 - 10 = 40 	ext{ cm}$.

$$A_{	ext{air}} = 50 	imes 40 	imes 10^{-4} 	ext{ m}^2 = 2000 	imes 10^{-4} 	ext{ m}^2$$

$$C_{	ext{air}} = \frac{1 \cdot \epsilon_0 \cdot (2000 	imes 10^{-4})}{5 	imes 10^{-2}} = 4\epsilon_0$$

Adding them together gives our total capacitance:

$$C_{eq} = 3\epsilon_0 + 4\epsilon_0 = 7\epsilon_0$$

Finally, we substitute the given value of $\epsilon_0 = 9 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}$:

$$C_{eq} = 7 	imes (9 	imes 10^{-12}) = 63 	imes 10^{-12} 	ext{ F}$$

Which is exactly **$63 	ext{ pF}$**. A perfectly elegant result!

Select Answer:

Visualized Solution

Geometry of Plates

Volume Formula

Substituting Values

Calculating Height $h$

Parallel Combination

Capacitance Formula

Setup for $C_{l i q u i d}$

Computing $C_{l i q u i d}$

Computing $C_{ai r}$

Equivalent Capacitance $C_{e q}$

Final Answer

The Way Forward

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Fluid Dynamics

The Master Equation

Final Calculation

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