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A capacitor of capacitance C is connected through a battery of emf E . when charging
is complete the capacitor is taken out from the circuit and then plate of the capacitor is cut
such that each plate is devided into two parts without
changing the distance between them and the areas of plates become A/3
and 2A/3 and then both capacitors are joined in series then what will be the heat that will produce.

Learn Everything , 8 Years ago
Grade 12
anser 1 Answers
Askiitians Tutor Team

To tackle this problem, we need to break it down into several steps, focusing on the behavior of capacitors and the energy transformations involved. Let's start by analyzing the initial conditions and then see how the changes affect the system.

Initial Charging of the Capacitor

When a capacitor of capacitance C is connected to a battery with an electromotive force (emf) E, it charges up to a voltage equal to E. The energy (U) stored in the capacitor can be calculated using the formula:

U = (1/2) C E²

Modifying the Capacitor Plates

After the capacitor is fully charged, it is disconnected from the battery. At this point, the charge (Q) on the capacitor is given by:

Q = C E

Now, the plates of the capacitor are cut into two parts, creating two new capacitors. The areas of the plates become A/3 and 2A/3, but the distance between the plates remains unchanged. The capacitance of a capacitor is given by:

C = ε₀ (A/d)

where ε₀ is the permittivity of free space and d is the distance between the plates. Therefore, the new capacitances can be calculated as follows:

  • For the first capacitor (area A/3): C₁ = ε₀ (A/3) / d = (ε₀ A) / (3d)
  • For the second capacitor (area 2A/3): C₂ = ε₀ (2A/3) / d = (2ε₀ A) / (3d)

Calculating the New Capacitances

Now, we can express the new capacitances in terms of the original capacitance C:

C₁ = C / 3

C₂ = (2/3) C

Connecting Capacitors in Series

When capacitors are connected in series, the total capacitance (C_total) can be calculated using the formula:

1/C_total = 1/C₁ + 1/C₂

Substituting the values we found:

1/C_total = 1/(C/3) + 1/((2/3)C)

This simplifies to:

1/C_total = 3/C + 3/(2C) = (6 + 3) / (2C) = 9/(2C)

Thus, the total capacitance is:

C_total = (2C) / 9

Energy Stored in the New Configuration

The charge on the new series combination of capacitors remains the same as the original charge Q, since they were disconnected from the battery. The energy stored in the new configuration can be calculated using:

U_total = (1/2) C_total Q²

Substituting the values:

U_total = (1/2) * (2C/9) * (C E)²

This simplifies to:

U_total = (C² E²) / 9

Calculating the Heat Produced

The heat produced (ΔU) when the capacitor is modified can be found by comparing the initial energy stored in the capacitor and the energy stored after the modification:

ΔU = U_initial - U_total

Substituting the values we derived:

ΔU = (1/2) C E² - (C² E²) / 9

To combine these, we need a common denominator:

ΔU = (4.5 C E²) / 9 - (C² E²) / 9 = (4.5 C E² - C² E²) / 9

Thus, the heat produced due to the modification of the capacitor plates is:

ΔU = (4.5 C E² - C² E²) / 9

In summary, the heat produced during this process results from the difference in energy stored before and after the capacitor plates were modified and connected in series. This illustrates how changes in physical configurations of capacitors can significantly affect their electrical properties and energy storage capabilities.

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