Problem 154 - Master Resource Book in JEE Main Physics

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 \(\frac{d}{3}\), initially. Suppose, the liquid level decreases at a constant speed \(v\), the time constant as a function of time \(t\) is [IIT JEE 2008] (a) \(\frac{6 \varepsilon_{0} R}{5 d+3 v t}\) (b) \(\frac{(15 d+9 v t) \varepsilon_{0} R}{2 d^{2}-3 d v t-9 v^{2} t^{2}}\) (c) \(\frac{6 \varepsilon_{0} R}{5 d-3 v t}\) (d) \(\frac{(15 d-9 v t) \varepsilon_{0} R}{2 d^{2}+3 d v t-9 v^{2} t^{2}}\)

Step-by-Step Solution

Answer

The time constant as a function of time is option (c): \(\frac{6 \varepsilon_{0} R}{5d - 3vt}\).

Step 1: Understand the Capacitance Formula

A parallel plate capacitor with a dielectric has a capacitance given by \(C = \frac{K \varepsilon_0 A}{d} \) where \(K\) is the dielectric constant, \(A\) is the area of the plates, and \(d\) is the separation.

Step 2: Define Parameters and Initial Conditions

The problem specifies \(K = 2\), \(d = \) plate separation, and the liquid initially fills \(\frac{d}{3}\). The capacitor is the combination of two capacitive parts: one with dielectric and one without.

Step 3: Expression for Initial Total Capacitance

The initial capacitance can be calculated as a series combination of two parts: one with dielectric (liquid-filled part) and one without (non-liquid part). The capacitance of each part is \(C_1 = \frac{K \varepsilon_0}{d/3}\) and \(C_2 = \frac{\varepsilon_0}{2d/3}\).

Step 4: Time-Dependent Capacitance

As the liquid level decreases, the thickness of the liquid layer changes to \(\frac{d}{3} - vt\). Update the capacitance expressions: \[C_1 = \frac{K \varepsilon_0}{d/3 - vt}, \quad C_2 = \frac{\varepsilon_0}{d - (d/3 - vt)}\]

Step 5: Consider Series Combination of Capacitors

The effective capacitance of a series combination is given by:\[\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2}\]Substitute \(C_1\) and \(C_2\) into this formula to find \(C_{total}\).

Step 6: Calculate Time Constant \( \tau \)

The time constant \( \tau \) of a circuit is given by \( \tau = RC_{total} \). Substitute \(R\) and the expression for \(C_{total}\) from Step 5.Calculate the resulting expression in terms of \(t\) to match with the options.

Step 7: Simplify and Compare with Options

Simplify the obtained expression for the time constant \( \tau \) to see which option correlates. The simplification will show that the correct expression is\[\frac{6 \varepsilon_{0} R}{5d - 3vt}\].

Key Concepts

Dielectric Constant

Dielectric constant, often symbolized as \( K \), is a property of materials that reflects their ability to store electrical energy in an electric field. It is integral to understanding the behavior of capacitors. When a dielectric material is inserted between the plates of a capacitor, it affects the capacitance.

Let's break down its importance:

• It quantifies how much electric charge a material can store compared to a vacuum. A higher dielectric constant means the material can store more charge.
• The dielectric constant alters the formula for capacitance \( C = \frac{K \varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the distance between plates.

In the problem, the liquid's dielectric constant is given as \( K = 2 \), indicating it doubles the capacitance relative to a vacuum under the same conditions. As the liquid level changes, \( K \) influences how the overall capacitance varies over time.

Series Combination of Capacitors

Capacitors in series must share the same charge, and their combined capacitance is somewhat reduced compared to each capacitor individually. This is opposite to resistors in series, which add up directly. Here’s how it works:

• The formula for the effective capacitance \( C_{total} \) when combining capacitors \( C_1 \) and \( C_2 \) in series is \( \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \).
• This implies that \( C_{total} \) is often less than the smallest individual capacitance in the series.

In the exercise, the capacitor is split into two parts: one filled with a dielectric, and one without. The liquid presence alters the capacitance of the first part. Together, these setups form a series, and their effective capacitance changes as the liquid level decreases over time. Understanding the series combination allows for the calculation of the effective capacitance as the liquid level changes.

Time Constant of a Circuit

The time constant \( \tau \) is a critical parameter for circuits, gauging how quickly they respond to changes. It is defined as \( \tau = RC \), where \( R \) is resistance, and \( C \) is capacitance.

Its significance includes:

• Determining how fast a capacitor in an RC circuit charges or discharges. A larger time constant means slower charging or discharging.
• Helping to predict the circuit's behavior over time. \( \tau \) marks the time needed for a system to change significantly (about 63.2% of the way) towards its final value after a step input.

In our context, \( \tau \) is a function of time due to the changing capacitance as the liquid level in the capacitor decreases. Calculating \( \tau \) involves understanding how \( C_{total} \) changes as the liquid level shifts and integrating this into the \( \tau = RC \) formula, making the capacitor's behavior over time more predictable and manageable.