Capacitors And Capacitance

1. Capacitance:

The capacitance of a capacitor is defined as the ratio of the charge stored to the potential difference across it:

From the graph, we can see that when , . Therefore:

2. Energy Stored:

The energy stored in a capacitor can be calculated using the following formula:

Using the values from the graph ( and ):

Alternatively, using the calculated capacitance:

The correct answer is (B) 12 μF, 600 μJ.

1. Number of capacitors in series:

To withstand the potential difference, we need to connect capacitors in series. The voltage across each capacitor in a series connection adds up to the total voltage. Therefore, the number of capacitors required in series is:

2. Equivalent capacitance of series combination:

The equivalent capacitance of capacitors in series is given by:

Since all the capacitors have the same capacitance ( ), the equivalent capacitance of the series combination is:

3. Number of parallel combinations:

To achieve the desired capacitance, we need to connect multiple series combinations in parallel. The equivalent capacitance of capacitors in parallel is the sum of their individual capacitances. Let be the number of parallel combinations required. Then:

4. Total number of capacitors:

The total number of capacitors needed is the product of the number of capacitors in series and the number of parallel combinations:

The correct answer is (D) 27.

1. Initial Capacitance:

The capacitance of a parallel plate capacitor without any dielectric is given by:

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

2. Capacitance with Dielectric Slab:

When a dielectric slab of thickness m and dielectric constant is introduced between the plates (without changing the separation ), the new capacitance becomes:

3. Capacitance with Increased Separation and Dielectric Slab:

To restore the capacitance to the original value , the separation between the plates is increased by m. The new separation is m. The capacitance with the increased separation and dielectric slab is :

4. Equating New and Original Capacitance:

We are given that the capacity is restored to the original value, so .

Therefore, we can equate the expressions for and :

5. Simplifying the Equation by Equating Denominators:

Since is present in the numerator of both sides and is non-zero, we can equate the denominators:

6. Solving for the Dielectric Constant :

Subtract from both sides of the equation:

Combine the numerical terms:

Add to both sides:

Divide both sides by :

Solve for :

Final Answer: The dielectric constant of the material will be 8.

A parallel plate capacitor with capacitance C₁ and a dielectric slab is connected to a battery, providing a potential difference V₁.

When the dielectric slab is removed while keeping the battery connected, the capacitance changes to C₂, and the new potential difference is V₂.

Understanding the Changes

The capacitance of a parallel plate capacitor with a dielectric is given by: where is the dielectric constant and is the capacitance without the dielectric.

When the dielectric is removed, the new capacitance becomes: Since , it follows that:

Effect on Voltage

Since the capacitor remains connected to the battery, the voltage across the plates is maintained at the same level:

Conclusion

From the above analysis: C₁ > C₂ and V₁ = V₂.

Hence, the correct answer is: V₁ = V₂, C₁ > C₂.

Given: Five capacitors each of value 1 μF are connected as shown in the figure. The equivalent capacitance between A and B needs to be calculated.

Step-by-Step Solution

Step 1: Identify the Configuration

We have two sets of capacitors in parallel connected to each other in series.

Step 2: Calculate Equivalent Capacitance of Each Set in Parallel

For the set connected to point A:

For the set connected to point B:

Step 3: Calculate Equivalent Capacitance of the Series Combination

Since the middle capacitor is in series with the parallel combinations, we calculate the equivalent capacitance as:

Step 4: Final Equivalent Capacitance

The final equivalent capacitance between points A and B is therefore: