Capacitors And Capacitance

The correct answer is (D) :
For a discharging capacitor when energy reduces to half the charge would become 1/√2 times the initial value.

Similarly,

To solve this problem, we need to understand the relationship between the length and resistance of wires based on the given conditions. Here's a step-by-step explanation:

1. The wire is 1 meter long, split into two parts and . Let be the length of the first part and be the length of the second part. Hence, .
2. The part is stretched into another wire such that the length of is twice that of . Therefore, the length of is .
3. The resistance of a wire is given by the formula , where is the resistivity, is the length, and is the cross-sectional area.
4. When the length of a wire is doubled (from to ), its cross-sectional area is halved to keep the volume of the wire constant, assuming uniform material. This means the resistance of becomes four times that of , i.e., .
5. According to the problem, the resistance of is twice that of , so . Thus, .
6. The resistance for wire is given as , and . From , we derive:

1. Simplifying, we get:

1. Assuming uniform area for both parts before stretching (i.e., before stretching), the equation simplifies to .
2. From and , solve for :

1. Substitute into :

1. The ratio of the length of to is:

Therefore, the correct answer is 1:2.

The correct answer is 10
At steady state potential difference across capacitor

Q = CV_c

The correct answer is 240


V = 20V





Therefore , the electrostatic energy of this system will be 240 ε₀ J

The correct answer is 8

⇒
⇒ d = 8 × 10^–3 m
⇒ x = 8

The circuit is equivalent to capacitors in parallel.

To solve the problem of finding the maximum current in an LC circuit formed by a capacitor and an inductor, we apply the principles of energy conservation. The energy stored in the capacitor is fully transferred to the inductor, reaching the maximum current. The energy initially stored in the capacitor is given by the formula: U = ¹⁄₂CV², where C = 500 μF = 500 × 10^-6 F and V = 100 V. Therefore, U = ¹⁄₂ × 500 × 10^-6 × (100)² = 2.5 joules.

In the LC circuit, this energy converts to magnetic energy stored in the inductor at maximum current, expressed as U = ¹⁄₂LI², where L = 50 mH = 50 × 10^-3 H. Solving for maximum current I gives 2.5 = ¹⁄₂ × 50 × 10^-3 × I². Simplifying, I²= 100 and thus I = √100 = 10 A.

The computed maximum current is 10 A, which neatly fits within the expected range of 10 to 10, confirming the solution's accuracy.

To find the potential difference between two identical thin metal plates with charges and , we consider the formation of a parallel plate capacitor. Given the situation, the potential difference between the plates is determined by the formula:

where:

• is the net charge on the plates.
• is the capacitance of the capacitor formed.

Initially, the charges on the two plates are and , respectively, with . When two plates are brought closer, they induce charges on each other and redistribute. The net charge that causes the potential difference is the difference in charges, that is:

However, due to the induction process, the effective potential difference caused by the charge difference is halved for a parallel plate capacitor. Therefore, the actual potential difference would be:

This expression matches with one of the provided options. Therefore, the correct answer is as follows:

Correct Answer:

Conclusion: The potential difference between the two plates, considering the induction effects in a parallel plate configuration, is given by the expression .

To solve the problem, follow these steps:

Initially, capacitor C₁ is charged to a potential of 30 V. The initial charge on C₁ is given by:

Q₁ = C₁ × V = 5 μF × 30 V = 150 μC

When the switch is closed, charge is conserved. The total charge in the system is shared between C₁ and C₂:

Total charge, Q_total = 150 μC

The potential difference across both capacitors at equilibrium is the same. Let V be this potential. Using charge conservation, we have:

(C₁ + C₂) × V = Q_total

V = 150 μC / (5 μF + 10 μF) = 150 μC / 15 μF = 10 V

Now, calculate the charge on C₂:

Q₂ = C₂ × V = 10 μF × 10 V = 100 μC

The charge on capacitor C₂ at equilibrium is 100 μC, which falls within the expected range of 100,100.

Thus, the charge on C₂ is 100 μC.

Given : Charge on C₁ = KCE
Charge on C₂ = CE
When they are connected in parallel charge will be equally divided so charge on one capacitor is



so, the correct answer is 55.

Step 1: Equivalent Capacitance of the Circuit

The total equivalent capacitance ( ) of the circuit is given as:

Step 2: Total Charge ( ) in the Circuit

The total charge stored in the circuit can be calculated using the formula:

where:

• (equivalent capacitance),
• (applied voltage).

Substitute the values:

Step 3: Charge Redistribution ( )

The charge on a specific branch of the circuit is calculated using the charge division formula. For the given branch:

Substitute the values:

• ,
• ,
• .

Simplify:

Final Answer:

To find the maximum current in the LC circuit, we can use the principle of conservation of energy. Initially, the energy is stored in the capacitor, and it is given by the formula:

where and .

Plugging in the values, we get:

In an LC circuit, the maximum energy stored in the capacitor (at its maximum charge) will be equal to the maximum energy stored in the inductor when the capacitor is fully discharged, i.e., when the current through the inductor is maximum. The energy in the inductor is given by:

Equating the energies, we have:

Given , we can solve for :

Thus, the maximum current in the circuit is 1.5 A, which matches the correct answer option: .

The current through a capacitor in an AC circuit is given by the formula:

where:

• is the supply voltage.
• is the capacitance.
• is the angular frequency.

The displacement current in the capacitor is the same as the conduction current in the circuit. The current through the capacitor is given by:

where is the capacitive reactance.

Now we calculate:

Thus, the rms value of the current is 13.8 for both the conduction and displacement current.

The problem asks for the supply voltage applied to a parallel plate capacitor that has been modified with dielectric materials. We are given the initial capacitance and dimensions, the dielectric constants of the new materials, and the resulting attractive force between the plates.

Concept Used:

The solution involves several key concepts from electrostatics:

1. Capacitors in Parallel: When the space between the plates of a capacitor is filled with two dielectric slabs side-by-side, each occupying half the area, the arrangement is equivalent to two separate capacitors connected in parallel.
2. Effect of Dielectrics on Capacitance: Inserting a dielectric of constant into a capacitor multiplies its original capacitance by . For a capacitor with area and separation , the capacitance of one part with dielectric and area is .
3. Equivalent Capacitance: The equivalent capacitance of a parallel combination is the sum of the individual capacitances: .
4. Force Between Capacitor Plates: The magnitude of the attractive force between the plates of a parallel plate capacitor is related to the energy stored and the plate separation by the formula . When the capacitor is connected to a voltage source , the stored energy is . Combining these gives the force as:

Step-by-Step Solution:

Step 1: List the given parameters and model the new capacitor configuration.

• Initial capacitance (with air), .
• Plate separation, .
• Dielectric constants, and .
• New force between the plates, .

The modified capacitor can be treated as two capacitors in parallel. Let be the capacitor with dielectric and area , and be the one with and area .

Step 2: Calculate the new equivalent capacitance .

The capacitance of the original air-filled capacitor is . The capacitances of the two new sections are:

The new equivalent capacitance is the sum of these two, as they are in parallel:

Step 3: Substitute the given values to find the numerical value of .

Step 4: Use the force formula to find the supply voltage .

The formula relating force, capacitance, voltage, and plate separation is:

We can rearrange this equation to solve for the voltage :

Step 5: Substitute the known values and calculate the final voltage.

The supply voltage is 80 V.

Step 1: Calculate the extra charge.

The change in charge ( ) due to the introduction of the dielectric is given by:



where K is the dielectric constant, C is the capacitance, and V is the voltage.









Step 2: Calculate the change in electrostatic energy.

The change in electrostatic energy ( ) is given by:

To determine the dimension of , we need to analyze the given expression for the position of a particle on the x-axis:

Let's discuss the terms one by one:

1. : This term involves the dimensionless function . Therefore, the dimension of must be to be consistent with , which has a dimension of length.
2. : Similarly, is dimensionless, so the dimension of is also .
3. : Here, has the dimension . Since must have the dimension of length, the dimension of is .
4. : As a constant term representing position, the dimension of is .

Now, we can determine the dimension of the expression :

Simplifying this, we get:

Therefore, the dimension of is .

The correct answer is therefore:

To solve this problem, let's analyze the circuit when the key is closed at the initial instant.

1. Understanding the Circuit:
   • The circuit consists of a resistor , a capacitor with plates and , a switch (key), and a 5V battery.
   • When the switch is closed, the capacitor will start charging from the battery.
2. Initial Instant Behavior:
   • At the very instant the switch is closed, the capacitor behaves like a short circuit because it is initially uncharged.
   • This means maximum current will flow through the wires as the capacitor starts to charge.
3. Analyzing each Statement:
   • Statement A: "There will be no current through resistor " - This is incorrect. At the instant of closing the key, maximum current flows through the circuit as the capacitor starts charging.
   • Statement B: "There will be maximum current in the connecting wires" - This is correct. The initial surge current is the maximum as the capacitor charges from zero volts.
   • Statement C: "Potential difference between the capacitor plates A and B is minimum" - This is correct. Initially, the potential difference is zero as the capacitor is uncharged.
   • Statement D: "Charge on the capacitor plates is minimum" - This is correct. At the beginning, the charge on the capacitor is zero.
4. Conclusion:
   • The correct combination of statements is B, C, and D, indicating valid observations at the instant the key is closed.