A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of di

Q: Why is the thickness of the dielectric $d/3 - vt$?

The problem states the initial level is $d/3$ and it decreases at a constant speed $v$. Distance equals speed multiplied by time, so the level drops by $vt$ after time $t$. Thus, the remaining thickness is the initial height minus the distance fallen: $d/3 - vt$.

Q: Can I treat this as two capacitors in series?

Yes, absolutely! A partially filled capacitor is equivalent to two capacitors in series: one filled with air (thickness $d - t'$) and one filled with the dielectric (thickness $t'$). Using $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$ will yield the exact same formula $C = \frac{\varepsilon_0 A}{d - t' + t'/K}$.

Q: What happens if the liquid completely drains out?

If the liquid completely drains, $t'$ becomes 0. The formula simplifies to $C = \frac{\varepsilon_0 A}{d}$, which is the standard capacitance for an air-filled parallel plate capacitor.

Q: Why did we use $A = 1$?

The problem explicitly states that the capacitor has 'plates of unit area'. In physics, 'unit' means a value of 1. So, we substitute $A = 1$ into our equations.

Q: Why does the time constant decrease as the liquid drains?

The liquid has a higher dielectric constant ($K=2$) than air ($K=1$). As it drains, it's replaced by air, which lowers the overall capacitance of the system. Since the time constant is $\tau = CR$, a smaller capacitance leads to a smaller time constant.

Question

A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K = 2$. The level of liquid is $d / 3$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is

Options:

(A) $\frac{6\varepsilon_0 R}{5d + 3vt}$

(B) $\frac{(15d + 9vt)\varepsilon_0 R}{2d^2 - 3dvt - 9v^2t^2}$

(C) $\frac{6\varepsilon_0 R}{5d - 3vt}$

(D) $\frac{(15d - 9vt)\varepsilon_0 R}{2d^2 + 3dvt - 9v^2t^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a parallel plate capacitor acting like a container, partially filled with a liquid dielectric. The liquid isn't static; it's draining out at a constant speed $v$. We are tasked with finding how the time constant of this RC circuit changes as the liquid level drops.

Initially, the liquid level is at $\frac{d}{3}$. Since it's decreasing at a constant speed $v$, the thickness of the liquid at any given time $t$ can be expressed as:
$$t' = \frac{d}{3} - vt$$

### The Master Equation
To find the time constant $	au = CR$, we first need to determine the equivalent capacitance $C$ of this partially filled capacitor. The general formula for a capacitor with plate area $A$, separation $d$, and a dielectric of thickness $t'$ and dielectric constant $K$ is:
$$C = \frac{\varepsilon_0 A}{d - t' + \frac{t'}{K}}$$

We are given that the plates have unit area ($A = 1$) and the dielectric constant of the liquid is $K = 2$. Let's substitute these values, along with our expression for $t'$, into the master equation:
$$C = \frac{\varepsilon_0 (1)}{d - \left(\frac{d}{3} - vtight) + \frac{\frac{d}{3} - vt}{2}}$$

### Simplifying the Denominator
This is where we need to be careful with our algebra. Let's simplify the denominator step-by-step. First, distribute the negative sign and split the fraction:
$$	ext{Denominator} = d - \frac{d}{3} + vt + \frac{d}{6} - \frac{vt}{2}$$

Now, group the terms with $d$ and the terms with $vt$:
$$	ext{Denominator} = \left(d - \frac{d}{3} + \frac{d}{6}ight) + \left(vt - \frac{vt}{2}ight)$$

Finding a common denominator for the $d$ terms (which is 6):
$$	ext{Denominator} = \left(\frac{6d - 2d + d}{6}ight) + \frac{vt}{2} = \frac{5d}{6} + \frac{vt}{2}$$

To make it a single fraction, multiply the second term by $3/3$:
$$	ext{Denominator} = \frac{5d + 3vt}{6}$$

### Final Calculation
Now, substitute this simplified denominator back into our capacitance equation. The 6 in the denominator of the fraction flips up to the numerator:
$$C = \frac{\varepsilon_0}{\frac{5d + 3vt}{6}} = \frac{6\varepsilon_0}{5d + 3vt}$$

Finally, the time constant $	au$ of an RC circuit is the product of its equivalent capacitance and resistance:
$$	au = CR = \left(\frac{6\varepsilon_0}{5d + 3vt}ight) R = \frac{6\varepsilon_0 R}{5d + 3vt}$$

**This perfectly matches option (a).** Notice how the time constant is inversely proportional to time $t$. As the liquid drains, the capacitance drops, and the circuit's response time becomes faster!

Select Answer:

Visualized Solution

Visualizing the Setup

Capacitance Formula

Substituting Known Values

Setting up the Equation

Simplifying the Denominator

Capacitance as a Function of Time

Calculating the Time Constant

Physical Interpretation

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup Imagine a parallel plate capacitor acting like a container, partially filled with a liquid dielectric

The Master Equation To find the time constant $τ = C R$ , we first need to determine the equivalent capacitance $C$ of this partially filled capacitor

Simplifying the Denominator This is where we need to be careful with our algebra

Final Calculation Now, substitute this simplified denominator back into our capacitance equation

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