Capacitance

1. Volume Conservation:

When eight small drops of mercury combine to form a big drop, the total volume of mercury remains conserved. Let be the radius of each small drop and be the radius of the big drop. The volume of a sphere is given by .

Volume of each small drop, .

Total volume of 8 small drops, .

Volume of the big drop, .

Since the volume is conserved, :

2. Relation between Radii:

Divide both sides of the volume equation by :

Take the cube root of both sides to find the relation between and :

So, the radius of the big drop is twice the radius of each small drop.

3. Capacitance of a Spherical Conductor:

The capacitance of a spherical conductor of radius is given by the formula , where is the permittivity of free space.

Capacitance of each small drop, .

Capacitance of the big drop, .

4. Comparing Capacitances:

Substitute into the expression for :

Since , we can write:

Final Answer: Thus, the capacitance of the bigger drop is 2 times the capacitance of each smaller drop.

Let's analyze the scenario of a capacitor being charged, disconnected, and then subjected to a dielectric material (glass slab) between its plates. Based on this analysis, we need to determine which pair of quantities decreases.

Step 1: Charging the Capacitor

The capacitor is charged by connecting a 2 Volt battery across it. Initially, the potential difference across the capacitor is 2 Volts. As the voltage is applied, the charge stored on the plates increases.

Step 2: Disconnecting the Battery

After disconnecting the battery, the potential difference across the capacitor remains the same as it was when charged, i.e., 2 Volts. The charge remains the same because there is no current flow, and the capacitor is isolated from the battery.

Step 3: Introducing the Glass Slab (Dielectric Material)

When a dielectric material, such as a glass slab, is introduced between the plates of the capacitor, it increases the capacitance of the capacitor. The capacitance with the dielectric is given by the formula:

Where:

• k: The dielectric constant of the material (for glass, typically )
• C₀: The original capacitance of the capacitor without the dielectric

As the capacitance increases due to the introduction of the glass slab, the charge stored on the plates also increases to maintain the same potential difference . This is because the relationship between charge, capacitance, and voltage is given by:

Conclusion:

Based on the analysis, we can conclude that the correct pair of quantities that decreases is:

• Potential difference and energy stored.

Correct answer: (D) Potential difference and energy stored.

The energy stored in a capacitor is given by:

where and .

The rate at which energy is stored in the capacitor is the derivative of the energy with respect to time:

Substituting the known values:

Thus, the rate at which energy is stored in the capacitor is 12 W.