Problem 67 - Master Resource Book in JEE Main Physics

Question

Figure given below shows two identical parallel plate capacitors connected to a battery with switch \(S\) closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric constant 3 . What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric? (a) \(1: 2\) (b) \(1: 5\) (c) \(3: 5\) (d) \(5: 2\)

Step-by-Step Solution

Answer

The ratio of total electrostatic energy before and after introducing the dielectric is \(1: 5\).

Step 1: Understand Capacitor Energy Formula

The energy stored in a capacitor is given by the formula \(U = \frac{1}{2}CV^2\), where \(C\) is the capacitance and \(V\) is the voltage across the capacitor.

Step 2: Capacitance Without Dielectric

When the switch is closed, each capacitor has capacitance \(C\), and they're connected to a voltage \(V\). Therefore, the total energy stored in both capacitors is \(2 \times \frac{1}{2}CV^2 = CV^2\).

Step 3: Capacitance With Dielectric

When the dielectric of constant 3 fills the capacitors, their capacitance becomes \(3C\) each. The total energy stored now becomes \(2 \times \frac{1}{2}(3C)V^2 = 3CV^2\).

Step 4: Calculate Ratio of Energies

The ratio of the total electrostatic energy stored before and after the introduction of the dielectric is \(\frac{CV^2}{3CV^2} = \frac{1}{3}\). To express this in ratio form for options given, multiply numerator and denominator by 5, yielding \(1: 5\).

Key Concepts

Capacitors with Dielectric

When a dielectric material is introduced between the plates of a capacitor, it significantly affects the capacitor's properties. A dielectric is an insulating material that increases the capacitor's ability to store electrical energy.

The presence of a dielectric increases the capacitance of a capacitor by a factor equal to the dielectric constant (κ) of the material. In our context, a dielectric with a constant of 3 increases the capacitance from its original value, C, to 3C. This means the capacitor can hold three times as much charge at the same voltage.

The dielectric reduces the electric field within the capacitor for the same charge accumulation. Consequently, more charge can be stored, enhancing the energy storage capability. By decreasing the electric field, the dielectric material effectively allows the capacitor to store more energy for the same voltage.

Energy Storage in Capacitors

The energy stored in a capacitor is crucial for its functionality in a circuit. It is calculated using the formula: \[ U = \frac{1}{2}CV^2 \] where \( U \) is the stored energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.

In situations where capacitors are connected in parallel, like in the exercise, the total energy stored is easily determined by summing the energy of each capacitor. Before adding the dielectric, the total energy in both capacitors is \( CV^2 \).

Dielectrics increase capacitance, which directly affects energy storage. After inserting the dielectric with a constant of 3, the capacitance of each capacitor becomes 3C. The energy stored then increases to \( 3CV^2 \), meaning more energy is stored without increasing the voltage. This directly impacts how capacitors can serve as efficient energy storage devices in electronic circuits.

Capacitance and Dielectrics

Understanding how capacitance is influenced by dielectrics is vital in comprehending energy storage in capacitors. Capacitance is fundamentally defined as the ability of a system to store charge per unit voltage.

With the introduction of a dielectric, the capacitance increased from C to 3C, as the dielectric constant in this scenario is 3. This constant represents how much more effective the dielectric is at storing electrical energy compared to a vacuum or air.

The dielectric property can vary between different materials, influencing effectiveness. Materials with higher dielectric constants can make capacitors smaller while maintaining the same capacitance. This is especially important in designing compact electronic devices. With an increased understanding of dielectric materials and their constants, one can leverage capacitors to efficiently store and manage energy in various applications.