Capacitance

Given:

In steady state, we have:

Step 1: Calculating the current:

The current is given by:

Step 2: Potential difference across capacitors:

The potential difference across capacitor is: The potential difference across capacitor is:

Step 3: Calculating the charge on the capacitors:

The charge on is: The charge on is:

Step 4: Ratio of the charges:

The ratio of the charges is:

The given problem involves finding the capacitance of a capacitor with a stair-like structure. Let's break it down step-by-step.

Understanding the Structure:

We have a flat plate of area and a second plate with a stair-like structure. The stair structure consists of three parts, each with an area and height .

Concept of Capacitance:

The capacitance of a parallel-plate capacitor is given by the formula:

where is the permittivity of free space, is the area of the plate, and is the separation between the plates.

Solution:

Each step of the stair functions as an individual capacitor in parallel with the others. The areas and distances are as follows:

1. First stair: Area = , Distance =
2. Second stair: Area = , Distance =
3. Third stair: Area = , Distance =

Each of these capacitors contributes to the total capacitance inversely proportional to their distance:

Capacitance for each stair:

Since these capacitors are in parallel, their total capacitance is the sum:

Substituting the values:

Taking a common denominator:

Conclusion:

The capacitance of the given arrangement is , matching the correct answer.

Given Circuit:

The circuit consists of a 10V battery, three resistors , , , and a capacitor .

In steady state: There will be no current in the branch with the capacitor, so there is no voltage drop across , and thus .

Step 1: The current in the rest of the circuit:

Step 2: The voltage drop across is given by:

Step 3: Since no current flows through the capacitor, the voltage across is equal to the voltage across the capacitor:

Step 4: Substituting the values:

Step 5: The charge on the capacitor is:

Final Answer: The charge on the capacitor is .

The problem asks for the value of , where the loss of energy upon connecting two capacitors is given as . Here, is the initial energy of the first capacitor.

Concept Used:

The solution is based on the principles of energy storage in capacitors and the conservation of charge.

1. Energy of a Capacitor: The electrical potential energy stored in a capacitor of capacitance charged to a potential is given by:

2. Conservation of Charge: When capacitors are connected, charge is redistributed, but the total charge of the isolated system is conserved. The total charge is the sum of the initial charges on the capacitors.

3. Common Potential: When capacitors are connected in parallel, they reach a common potential , which is calculated as:

4. Energy Loss: The loss of energy is the difference between the total initial energy and the total final energy of the system. This loss is dissipated as heat in the connecting wires.

Step-by-Step Solution:

Step 1: Calculate the initial energy of the system.

The system initially consists of two separate capacitors.

• For the first capacitor: Capacitance , Potential . Its energy is . The problem states that this is .
• For the second capacitor: Capacitance , Potential . Its energy is .

The total initial energy is the sum of the energies of the two capacitors:

Step 2: Calculate the common potential after connecting the capacitors.

First, find the initial charge on each capacitor:

When connected in parallel (assuming positive plates are connected together), the total charge is conserved:

The total capacitance of the parallel combination is:

The common potential is:

Step 3: Calculate the final energy of the system.

The final energy is the energy stored in the equivalent capacitor at the common potential:

Step 4: Calculate the loss of energy.

To subtract, we find a common denominator:

Final Computation & Result:

Step 5: Express the energy loss in terms of and find the value of .

We know that the initial energy of the first capacitor is .

We can rewrite the energy loss in terms of :

The problem states that the loss of energy is . Comparing this with our result:

This implies that .

The value of is 2.

To solve this problem, we need to understand how the capacitance of a parallel plate capacitor can be increased. The capacitance is given by the formula:v

where:

• is the capacitance.
• is the permittivity of free space (a constant).
• is the area of one of the plates, computed as .
• is the distance between the plates.

Initially, the values given are:

The area is:

Therefore, the initial capacitance is:

To increase the capacitance by a factor of 10, the new capacitance should be:

We need to identify which options achieve this new capacitance by either increasing the plate area or reducing the distance between the plates:

1. Option A:
2. Option B:
3. Option C:
4. Option D:
5. Option E:

Thus, the correct choices are C and E only.

To find the potential energy stored in a parallel plate capacitor, we start by considering the formula for the energy stored in a capacitor. The energy stored in a capacitor is given by:

where is the capacitance of the capacitor and is the potential difference across the plates.

The capacitance of a parallel plate capacitor is given by:

where is the permittivity of free space, is the area of the plates, and is the separation between the plates.

The electric field between the plates of the capacitor is given by:

Substituting in the energy formula, we have:

Substituting the expression for , we get:

Upon simplifying, this expression becomes:

This matches the given option , confirming it as the correct choice.