Problem 28 - Master Resource Book in JEE Main Physics

Question

A slab of copper of thickness, \(b\) is inserted in between the plates of parallel plate capacitor as shown in figure. The separation between the plates is \(d\) if \(b=d / 2\), then the ratio of capacities of capacitors after and before inserting the slab will be (a) \(\sqrt{2}: 1\) (b) \(2: 1\) (c) \(1: 1\) (d) \(1: \sqrt{2}\)

Step-by-Step Solution

Answer

(c) \(1: 1\)

Step 1: Understand the Initial Configuration

Initially, without the copper slab, the capacitor consists of two plates separated by a distance \(d\). The initial capacitance \(C_0\) is given by the formula: \[ C_0 = \frac{\varepsilon_0 A}{d} \] where \(\varepsilon_0\) is the permittivity of free space and \(A\) is the area of the plates.

Step 2: Introduce the Copper Slab

When a copper slab of thickness \(b = \frac{d}{2}\) is inserted between the plates, it divides the capacitor into two smaller capacitors in series. The slab has no potential difference across it since it is a conductor, so the effective separation between the two resulting capacitors is \(\frac{d}{2}\) on each side of the slab.

Step 3: Calculate Capacitors in Series After Insertion

Each part of the capacitor now has a separation of \(\frac{d}{2}\), and the capacitance \(C_1\) for each of these sections is given by \[ C_1 = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2\varepsilon_0 A}{d} \]. Since these capacitors are in series, the total capacitance \(C\) of the system after insertion is given by \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_1} = \frac{2}{C_1} \], thus, \[ C = \frac{C_1}{2} = \frac{\varepsilon_0 A}{d} \].

Step 4: Find the Ratio of Capacities

The capacitance after inserting the slab \(C\) is the same as the initial capacitance \(C_0\) because \[ C = \frac{\varepsilon_0 A}{d} = C_0 \]. Hence, the ratio of capacitance after to before the insertion is \(\frac{C}{C_0} = \frac{C_0}{C_0} = 1:1\).

Key Concepts

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by a distance, with an insulating material, known as a dielectric, in between. It is an essential component in storing electrical energy. The capacitance of a parallel plate capacitor is calculated using the formula:

• \[ C_0 = \frac{\varepsilon_0 A}{d} \]

Here:

• \( \varepsilon_0 \) is the permittivity of free space, a constant.
• \( A \) is the area of the plates.
• \( d \) is the distance between the plates.

The larger the area or the closer the plates, the greater the capacitance, meaning more energy can be stored. The separation plays a crucial role in determining how much charge the capacitor can store at a given voltage.
A parallel plate capacitor is practical, as it is straightforward in construction and understanding.

Copper Slab Insertion

Inserting a copper slab between the capacitor plates adjusts the capacitor's configuration. Copper, being a conductor, doesn't allow any potential difference across its thickness, altering how the capacitor operates. In our exercise, the copper slab is half the thickness of the total plate separation (\( b = \frac{d}{2} \)).
Placing it into the capacitor creates two new capacitors with a separation of \( \frac{d}{2} \) on either side of the slab.

• The copper slab doesn't affect the electric field.
• It divides the overall capacitor into two sections, effectively acting as capacitors in parallel in terms of space but are in series electrically.

This alteration doesn’t change the total capacitance of the system. Each new segment has a capacitance of:

• \[ C_1 = \frac{2\varepsilon_0 A}{d} \]

Calculating these as capacitors in series can be tricky at first but is vital in understanding the final configuration.

Series Capacitors

When capacitors are connected in series, the total effect is different compared to when they are in parallel. Rather than simply adding up their individual capacitances, the inverse of the total capacitance is the sum of the inverses of each capacitor's capacitance.
For two capacitors in series, the formula is:

• \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_1} \]

This simplifies to:

• \[ C = \frac{C_1}{2} \]

The exercise shows that, post-slab insertion, the total capacitance becomes equivalent to the initial capacitance of a single, unchanged capacitor. This result might seem counterintuitive but is a hallmark of capacitors in series: when they're divided equally by a perfect conductor in this manner, no changes occur in total capacitance.
Understanding series capacitors helps in predicting the behavior of complex circuits and highlights an essential principle in electric circuitry and energy storage.