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Chapter 17: Problem 52

An air capacitor \(C\) connected to a battery of \(\operatorname{emf} V\) acquires a charge \(q\) and energy \(E\). The capacitor \(C\) is disconnected from the battery and a dielectric slab is placed between the plates. Which of the following statements is correct? (a) \(V\) and \(q\) decreases but \(C\) and \(E\) increases. (b) \(V\) remains unchanged, but \(q, E\) and \(C\) increases. (c) \(q\) remains unchanged, \(C\) increases, \(V\) and \(E\) decreases. (d) \(q\) and \(C\) increases but \(C\) and \(E\) decreases.

Short Answer

Expert verified
(c) \(q\) remains unchanged, \(C\) increases, \(V\) and \(E\) decrease.

Step by step solution

01

Initial Observation

When a capacitor is connected to a battery, it is charged up to voltage \(V\) and acquires charge \(q\). The capacitance \(C\) is defined by \(C = \frac{q}{V}\) and energy \(E\) is given by \(E = \frac{1}{2} C V^2\).
02

Disconnecting the Battery

Once the capacitor is disconnected from the battery, the charge \(q\) on the capacitor remains constant because there is no path for the charge to leave or change. The voltage \(V\) across the capacitor is determined by the charge and the capacitance: \(V = \frac{q}{C}\).
03

Inserting the Dielectric

Inserting a dielectric increases the capacitance of the capacitor because capacitance is given by \(C' = \kappa C\), where \(\kappa\) is the dielectric constant. This means the capacitance \(C\) increases when the dielectric is placed between the plates.
04

Effect on Voltage

Since the charge \(q\) remains constant, the new voltage \(V'\) after inserting the dielectric is given by \(V' = \frac{q}{C'} = \frac{q}{\kappa C}\), which means the voltage decreases.
05

Effect on Energy

The energy stored in a capacitor is given by \(E = \frac{1}{2} \frac{q^2}{C}\). After inserting the dielectric, the new energy \(E'\) becomes \(E' = \frac{1}{2} \frac{q^2}{C'} = \frac{1}{2} \frac{q^2}{\kappa C}\), which shows that the energy decreases due to the increase in capacitance.

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Key Concepts

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

Dielectric Material
A dielectric material is a non-conducting substance introduced between the plates of a capacitor. Its primary function is to enhance the capacitor's storage capacity. When you place a dielectric slab between the plates, it boosts the capacitor's capacitance by a factor equivalent to the dielectric constant, denoted by \( \kappa \).
This constant is a measure of the material's ability to resist electric fields, effectively allowing the capacitor to hold more charge at the same voltage.
  • The capacitance with the dielectric becomes \( C' = \kappa C \).
  • The dielectric does not allow free flow of charge but instead polarizes, increasing the effective field across the plates.
This results in increased capacitance without needing to change physical dimensions or charges.
Voltage
Voltage is the electric potential difference between two points. For capacitors, it's the potential difference between the plates. When connected to a battery, the capacitor charges up to a voltage \( V \). Upon disconnecting the battery, the charge \( q \) on the capacitor remains constant.
  • If a dielectric is inserted, the voltage across the capacitor changes.
  • The new voltage \( V' \) is lower: \( V' = \frac{q}{C'} = \frac{q}{\kappa C} \).
This decrease happens because while the charge stays the same, the capacitance increases due to the dielectric. Ultimately, the drop in voltage aligns with the equation \( V = \frac{q}{C} \), adjusting for the increased capacitance.
Energy in Capacitor
The energy stored in a capacitor originates from the electric field between its plates and can be calculated using different formulas, including \( E = \frac{1}{2} C V^2 \) or \( E = \frac{1}{2} \frac{q^2}{C} \).
When we introduce a dielectric material, the capacitance increases to \( C' = \kappa C \), and the energy also transforms.
  • The new energy \( E' \) can be determined as \( E' = \frac{1}{2} \frac{q^2}{C'} = \frac{1}{2} \frac{q^2}{\kappa C} \).
By increasing the capacitance with the dielectric, the energy stored in the capacitor decreases because \( E' = \frac{1}{2} \frac{q^2}{C'} \), showing that for a constant charge, a higher capacitance means less energy is needed.
Charge Conservation
Charge conservation is a principle stating that the total electric charge in an isolated system remains constant over time. In the context of capacitors, once a capacitor is disconnected from an external source like a battery, the charge held by the capacitor becomes conserved and cannot change because no path exists for it to move or exchange.
  • The charge \( q \) remains unchanged when a dielectric is inserted between the plates.
  • This results in modifications to other properties, but the amount of charge stays fixed.
Given that charge conservation holds, any induced changes such as with capacitance or voltage adjustments follow this principle, ensuring the net charge doesn't alter even as the system dynamics do.

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Most popular questions from this chapter

A variable parallel plate capacitor and an electroscope are connected in parallel to a battery. The reading of the electroscope would be decreased by (i) increasing the area of overlap of the plates (ii) placing a block of paraffin wax between the plates (iii) decreasing the distance between the plates (iv) decreasing the battery potential Then state if (a) only (i), (ii) and (iii) are correct (b) only (i) and (iii) are correct (c) only (ii) and (iv) are correct (d) only (iv) is correct

A parallel plate capacitor is charged. If the plates are pulled apart (a) the capacitance increases (b) the potential difference increases (c) the total charge increases (d) the charge and potential difference remain the same

1000 small water drops each of radius \(r\) and charge \(q\) coalesce together to form one spherical drop. The potential of the bigger drop is larger than that of the smaller one by a factor (a) 1000 (b) 100 (c) 10 (d) 1

Two identical capacitors are joined in parallel, charged to a potential \(V\), separated and then, connected in series, i.e., the positive plate of one is connected to the negative plate of the other. Then (a) the charges on the free plates connected together are destroyed (b) charges on the free plates are destroyed (c) the energy stored in the system increases (d) the potential difference between the free plates is \(2 \mathrm{~V}\)

There is an air-filled \(1 \mathrm{pF}\) parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to \(2 \mathrm{pF}\). The dielectric constant of wax is (a) 2 (b) 4 (c) 6 (d) 8
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