Problem 155 - Master Resource Book in JEE Main Physics

Question

Two condensers of capacities \(2 C\) and \(C\) are joined in parallel and charged upto potential \(V\). The battery is removed and the condenser of capacity \(C\) is filled completely with a medium of dielectric constant \(K\). The potential difference across the capacitors will now be \(\quad\) [IIT 1988; Similar AMU (Engg 2009] (a) \(\frac{3 V}{K+2}\) (b) \(\frac{3 \mathrm{~V}}{\mathrm{~K}}\) (c) \(\frac{V}{K+2}\) (d) \(\frac{V}{K}\)

Step-by-Step Solution

Answer

The potential difference across the capacitors will be \( \frac{3V}{2 + K} \).

Step 1: Initial Charge Calculation

First, calculate the total capacitance when the two capacitors are in parallel. The total capacitance is the sum of the individual capacitances: \[ C_{total} = 2C + C = 3C \]Charging them at potential \(V\), the initial total charge \(Q\) on the system is given by:\[ Q = C_{total} \times V = 3C \times V \]

Step 2: Effect of Dielectric

Once the battery is removed, the charge remains constant. Now, place a dielectric of constant \(K\) in the capacitor with capacitance \(C\). This causes its new capacitance \(C'\) to become:\[ C' = KC \]The capacitance of the other capacitor \(2C\) remains unchanged.

Step 3: Re-calculate the Total Capacitance

After inserting the dielectric, calculate the new total capacitance:\[ C_{new ext{-}total} = 2C + KC \]

Step 4: New Potential Difference Calculation

As charge \(Q\) is conserved and remains the same, use the formula for potential difference, \(V_{new}\), across the system:\[ V_{new} = \frac{Q}{C_{new ext{-}total}} = \frac{3CV}{2C + KC} \]To simplify, we divide both the numerator and the denominator by \(C\):\[ V_{new} = \frac{3V}{2 + K} \]

Key Concepts

Dielectric Constant

In the world of capacitors, the dielectric constant is an important concept. It describes how a medium, such as a dielectric, affects a capacitor's electrical properties.
When a dielectric is inserted between the plates of a capacitor, it increases the capacitor's ability to store charge. This happens because the dielectric reduces the electric field within the capacitor. This reduction effectively increases the capacitance.
So, what's the math behind this? The dielectric constant, denoted by \(K\), is a measure of this effect. It is defined as the ratio of the capacitance of a capacitor with the dielectric \( (C')\) to its original capacitance \((C)\):

• \( C' = KC \)

A higher dielectric constant means the medium is better at increasing capacitance. In many practical applications, understanding and utilizing materials with different dielectric constants is crucial for designing efficient capacitors.

Potential Difference

Potential difference, often referred simply as voltage, is a fundamental concept in electricity. It represents the work needed to move a charge between two points in an electric field. In a capacitor, this refers to the work required to move charges between the two plates.
When capacitors are joined in parallel, they experience the same potential difference across all components. In our exercise example, the potential difference is denoted by \(V\). At first, both capacitors are charged to this common voltage. When one capacitor is filled with a dielectric, the potential difference changes.
Even though the charge remains unchanged, the introduction of the dielectric alters the capacitance. Because of this, the new potential difference \(V_{new}\) must be recalculated using the new total capacitance \(C_{\text{new-total}}\):

• \( V_{new} = \frac{Q}{C_{\text{new-total}}} \)

This understanding helps in predicting how the voltage across capacitors will adjust when a dielectric is introduced, ensuring the conservation of charge in the system.

Charge Conservation

Charge conservation is one of the core principles of electromagnetic theory. It states that the total electric charge in an isolated system remains constant.
For capacitors, this principle means that once they are charged and disconnected from a battery, their total charge doesn't change, even if changes occur in the system, such as the introduction of a dielectric.
In our example, the initial charge \(Q\) was calculated when the capacitors were first charged:

• \( Q = C_{\text{total}} \times V \)

After disconnecting the battery and adding a dielectric to one capacitor, the charge across the capacitors must still total this original \(Q\). This continuity helps determine the new potential difference.
The conservation of charge ensures that calculations for new conditions, such as different capacitances or materials, remain consistent and predictable.