Capacitors And Capacitance

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by

Step 1: Energy formula of capacitor. U = (1)/(2)CV²
Step 2: Calculating energies. Initial energy: U₁ = (1)/(2) × 6×10⁻⁶ × 10² Final energy: U₂ = (1)/(2) × 6×10⁻⁶ × 20²
Step 3: Increase in energy. Δ U = U₂ - U₁ = 9 × 10⁻⁴ J

Step 1: Capacitors in parallel.
When capacitors are connected together, the charge redistributes such that the potential on each becomes zero. The relation between the capacitors is found by equating the charge on each.

Step 2: Apply charge conservation.
The condition for potential to become zero is . Thus, the correct answer is (2).

Step 1: Understand the energy stored in a capacitor.
The energy stored in a capacitor is .
Step 2: Calculate the total energy before and after connection.
Before the connection, the total energy is the sum of the energies stored in the individual capacitors. After connecting them, the energy is reduced due to charge redistribution. The decrease in energy is . Final Answer:

For Initial
For Final

Step 1: Capacitors and are in series:
Step 2: This combination is in parallel with :

Step 1: Initial charges:
Step 2: Since connected oppositely, net charge:
Step 3: Equivalent capacitance:
Step 4: Final voltage:
Step 5: Final energy:

Step 1: In steady state, capacitors act as open circuits.
Step 2: Voltage divides equally across equal resistances.
Step 3: Total voltage =100V: VAB = VBC = 50V

Step 1: The capacitor behaves as two capacitors in parallel.
Step 2: Capacitances: C₁ = (Kvarepsilon₀(A/2))/(d/2), C₂ = (Kvarepsilon₀(A/2))/(d)
Step 3: Total capacitance: C = C₁ + C₂ = (3Kvarepsilon₀A)/(4d)

Step 1: When connected by wire, both shells are at same potential.
Step 2: Equivalent capacitance is sum of individual capacitances. C = 4πvarepsilon₀ a + 4πvarepsilon₀ b

Step 1: Each vertical column acts as parallel combination.
Step 2: The infinite ladder reduces to a finite equivalent by symmetry.
Step 3: Solving recursively gives Cₑq=(4)/(3)μF

Explanation:
When two conducting shells are connected by a conducting wire, they form an equipotential system. This means that the potential difference ( ) between the two shells is zero because the wire ensures that both shells are at the same electric potential.
Let's denote the radii of the inner and outer shells as and respectively, with .
Step-by-Step Analysis:
1. Potential on the shells:
Since the shells are connected by a wire, the potential on both shells must be the same, say .
2. Charge Distribution:
- Let's assume that the inner shell has a charge .
- By electrostatic induction, the inner surface of the outer shell will have a charge , and the outer surface of the outer shell will have a charge to maintain neutrality.
3. Electric Potential Calculation:
- The potential on the inner shell of radius due to its own charge is:

- The potential on the outer shell of radius due to the charge on it is:

Since the potentials are the same ( ), we have:

This equation implies that for , to maintain equality, since there cannot be any potential difference.
4. Capacitance Calculation:
- Capacitance is defined as the charge stored per unit potential difference :

- Here, as the shells are at the same potential. In electrostatics, the capacitance of such a system is effectively infinite because the shells can theoretically hold an unlimited amount of charge without creating any potential difference.

Conclusion:
The capacity (capacitance) of the system, when two conducting shells are connected by a wire and made to be at the same potential, is theoretically infinite. This is because the potential difference is zero, leading to an infinite capacitance value mathematically.
So, the final impressive and detailed answer is:
When two conducting shells of radii and are connected by a conducting wire, they form an equipotential system. The potential difference between them is zero ( ). Thus, the capacitance of the system is given by:

Therefore, the capacitance of the system is theoretically infinite i.e correct Answer is Option 4 .

Step 1: Initial charges: Q₁ = CV, Q₂ = 2C(2V)=4CV
Step 2: Since opposite terminals are connected, net charge: Qₙet = 4CV - CV = 3CV
Step 3: Equivalent capacitance: Cₑq = C + 2C = 3C
Step 4: Final voltage: Vf = fracQₙetCₑq = V
Step 5: Final energy: U = (1)/(2)CₑqVf² = (1)/(2)(3C)V² = (3)/(2)CV²

Step 1: In steady state, capacitors act as open circuits and no current flows through them.
Step 2: Only the resistive path remains active. The resistors of 20Ω and 10Ω are in series with a 100V battery.
Step 3: Total resistance: Rₑq = 20 + 10 = 30Ω
Step 4: Current in the circuit: I = (100)/(30) = (10)/(3)A
Step 5: Potential drops: VAB = I × 20 = (10)/(3)×20 = 50V VBC = I × 10 = (10)/(3)×10 = 50V

Step 1: Left half of area (A)/(2) has two dielectrics in series: C₁=((d/2)/(Kvarepsilon₀(A/2))+(d/2)/(varepsilon₀(A/2)))⁻1 =(Kvarepsilon₀A)/((K+1)d)
Step 2: Right half of area (A)/(2) filled with dielectric K: C₂=(Kvarepsilon₀(A/2))/(d)
Step 3: Both parts are in parallel: C=C₁+C₂ =(varepsilon₀A)/(d)((K)/(K+1)+(K)/(2))
Step 4: Simplifying: C=(K(K+3)varepsilon₀A)/(2(K+1)d)

To solve the problem of finding the potential difference across capacitor , follow these steps:

Step 1: Find the equivalent capacitance for capacitors in series using the formula:
$ Q = C \cdot V Q C_{eq} V_{total} C_2 Q = C \cdot V V C_2$ is 24 V.