/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 Two identical capacitors \(A\) a... [FREE SOLUTION] | 91Ó°ÊÓ

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Two identical capacitors \(A\) and \(B\) are joined in parallel to a battery. If a dielectric slab of constant \(K\) is slipped between the plates of capacitor \(B\) and battery remains connected then the energy of capacitor \(A\) will : (a) decrease (b) increase (c) remains the same (d) first increase then will again come to original value after process is completed

Short Answer

Expert verified
(c) remains the same

Step by step solution

01

Identifying Initial Conditions

Initially, capacitors \(A\) and \(B\) are joined in parallel to a battery, which implies they both have the same potential difference \(V\) across them. The energy stored in each capacitor can be described by the formula \( U = \frac{1}{2}CV^2 \), where \(C\) is the capacitance of each capacitor.
02

Introducing Dielectric to Capacitor B

When the dielectric material of constant \(K\) is inserted into capacitor \(B\), its capacitance becomes \(C' = KC\). This change does not affect the potential \(V\) across the capacitors since they remain connected to the battery.
03

Impact on Capacitor A

Since capacitors are connected in parallel to the battery, the potential difference \(V\) remains constant for both capacitors. The energy in capacitor \(A\) is still \( U_A = \frac{1}{2}CV^2 \). There is no change in its capacitance or potential, thus the energy remains unchanged.
04

Conclusion about Energy

The energy stored in capacitor \(A\) does not change when a dielectric is added to capacitor \(B\) because the potential across \(A\) remains constant, and its capacitance has not changed. Therefore, the energy of capacitor \(A\) remains the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel Circuits
In a parallel circuit, components are connected across the same voltage source, maintaining a consistent potential difference across each component. This is a key characteristic when dealing with capacitors connected in parallel.

The defining feature of parallel circuits is:
  • Same voltage across each component: In the given exercise, capacitors \( A \) and \( B \) both experience the same potential difference \( V \) due to their connection to a single battery source.
  • Independent component operation: Each capacitor's capacitance and the resulting stored energy can be calculated independently, without affecting other components in the circuit.
In practical applications, parallel circuits ensure that when one component undergoes a change, such as the insertion of a dielectric in one capacitor, it doesn't affect the voltage or energy across the others, as seen with capacitor \( A \). The energy stored in a component only changes if its capacitance or the circuit's overall potential difference changes, neither of which happens to capacitor \( A \) in this scenario.
Capacitance
Capacitance is a property of a capacitor that indicates its ability to store charge when a potential difference exists across its plates. It is defined as the ratio of the charge \( Q \) on each conductor to the voltage \( V \) between them, expressed as \( C = \frac{Q}{V} \). In the case of the capacitors \( A \) and \( B \), both initially have the same capacitance due to their identical nature.

When a dielectric material is introduced between the plates of a capacitor, its capacitance increases by a factor of the dielectric constant \( K \).
  • This is mathematically represented as: \( C' = KC \), where \( C' \) is the new capacitance.
  • For capacitor \( B \), the introduction of the dielectric affects its capacitance, making it \( K \) times greater than its initial value.
This concept is crucial because it helps explain why the energy in capacitor \( A \) remains unchanged. Even with the modification in capacitor \( B \), the unchanged capacitance and potential difference in capacitor \( A \) means its stored energy does not fluctuate.
Electric Potential
Electric potential refers to the amount of work needed to move a unit charge from a reference point to a specific point inside the field without producing an acceleration. In the context of capacitors in a circuit, it highlights the potential difference or voltage across the capacitor plates.

For capacitors in parallel, like \( A \) and \( B \):
  • The potential difference \( V \) remains constant regardless of changes within the circuit, unless additional modifications to the circuit or connections occur.
  • In this exercise, the battery maintains this constant voltage, ensuring that all capacitors in the parallel circuit experience the same electric potential.
This is fundamental because the energy of a capacitor is heavily dependent on both the capacitance \( C \) and the potential \( V \) as \( U = \frac{1}{2}CV^2 \). For capacitor \( A \), since both its capacitance and the potential difference remain unchanged throughout the process, the energy stored within it remains constant, demonstrating the lack of influence due to changes in an adjacent component.

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Most popular questions from this chapter

A capacitor of capacitance \(C_{0}\) is charged to a potential \(V_{0}\) and then isolated. A small capacitor \(C\) is then charged from \(C_{0}\), discharged and charged again, the process being repeated \(n\) times. Due to this, potential of the larger capacitor is decreased to \(V\). Value of \(C\) is : (a) \(C_{0}\left(\frac{V_{0}}{V}\right)^{1 / n}\) (b) \(C_{C}\left[\left(\frac{V_{0}}{V}\right)^{1 / n}-1\right]\) (c) \(C_{\mathrm{C}}\left[\left(\frac{V_{0}}{V}\right)-1\right]^{n}\) (d) \(C_{0}\left[\left(\frac{V}{V_{0}}\right)^{n}+1\right]\)

An isolated metallic object is charged in vacuum to a potential \(V_{0}\), its electrostatic energy being \(W_{0}\). It is then disconnected from the source of potential, its charge being left unchanged and is immersed in a large volume of dielectric, with dielectric constant \(K\). The electrostatic energy will be (a) \(K W_{0}\) (b) \(W_{0} / K\) (c) \(\frac{W_{0}}{2 K}\) (d) \(W_{0}\)

A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles, then: (a) the voltage across the plate increases (b) the capacitance increases (c) the electrostatic energy stored in the capacitor increases (d) both (a) and (c) are correct

A dielectric slab of thickness \(d\) is inserted in a parallel plate capacitor whose negative plate is at \(x=0\) and positive plate is at \(x=3 d\). The slab is equidistant from the plates. The capacitor is given some charge, as \(x\) goes from 0 to \(3 d\), then : (a) the direction of the electric field remains the same (b) the electric potential increases continuously (c) the electric potential increases at first, then decreases and again increases (d) both (a) and (b) are correct

Inside two identical capacitors, two identical dielectric slabs are introduced as shown in figure. What will happen, if slab of capacitor \(B\) is pulled out, with the battery remain connected ? (a) During the process charge flows from \(a\) to \(b\) (b) Finally charge on \(B\) will be less than charge on \(A\) (c) During the process work done by external force \(F\) appear as heat in the circuit (d) None of the above

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