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Chapter 25: Problem 81

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) \(W_{0}\)

Short Answer

Expert verified
The electrostatic energy will be \(\frac{W_{0}}{K}\).

Step by step solution

01

Understand the Initial Conditions

The problem first describes an isolated metallic object charged in a vacuum to a potential \(V_{0}\) with electrostatic energy \(W_{0}\). The object is disconnected from the source, which means the charge on the object is constant. The initial energy formulation is based on this constant charge.
02

Energy Formula for a Metal Object

The electrostatic energy \(W\) of a charged conductor is given by the formula \(W = \frac{1}{2} QV\), where \(Q\) is the charge and \(V\) is the potential. Initially, we denote the energy as \(W_{0} = \frac{1}{2} Q V_{0}\).
03

Effect of Dielectric on Potential

When a dielectric is introduced around the metallic object, the potential \(V\) changes. The potential of a charged object immersed in a dielectric reduces to \(V = \frac{V_{0}}{K}\), because dielectric constant \(K\) reduces the electric field and potential by a factor of \(K\).
04

New Electrostatic Energy in Dielectric

The new electrostatic energy \(W_d\) of the metallic object in the dielectric is \(W_d = \frac{1}{2} Q \cdot \frac{V_{0}}{K} = \frac{1}{2} \frac{QV_{0}}{K}\). Since \(W_{0} = \frac{1}{2} Q V_{0}\), substituting gives us \(W_d = \frac{W_{0}}{K}\).
05

Choose the Correct Answer

From the calculated energy, \(W_d = \frac{W_{0}}{K}\), the correct option among the given choices is (b) \(\frac{W_{0}}{K}\).

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Key Concepts

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

Dielectric Constant
The dielectric constant, represented as \( K \), is a measure of a material's ability to reduce the electric field and potential around a charged object. When a dielectric material is placed around a conductor, it affects how the electric field behaves. The dielectric constant quantifies this effect.
It is defined as the ratio of the electric permeability of the material to the electric permeability of free space.
  • A higher dielectric constant means the material can significantly weaken the electric field.
  • When using a dielectric with a conductor, the potential decreases because the electric field is weaker.
Therefore, placing a dielectric around a charged metallic object reduces the potential \( V \) to \( \frac{V_{0}}{K} \), effectively lowering the energy stored in the conductor.
Electrostatic Energy
Electrostatic energy refers to the potential energy stored in a charged conductor due to its position within an electric field. This energy is calculated using the formula \( W = \frac{1}{2} QV \), where \( Q \) is the charge and \( V \) is the electric potential.
  • Initially, the energy is \( W_{0} = \frac{1}{2} Q V_{0} \) in a vacuum.
  • When a dielectric is introduced, the energy becomes \( W_{d} = \frac{1}{2} \frac{Q V_{0}}{K} \).
This shows that electrostatic energy depends directly on both the charge and potential of the conductor. When potential decreases due to the presence of a dielectric, so does the energy stored in it.
Charged Conductor
A charged conductor is a material that can conduct electricity and be charged by a source. When charged, it holds electrostatic energy proportional to its potential and accumulated charge.
The conductor's charge remains constant once disconnected from the source, as stipulated in the exercise.
  • This property means any changes in energy or potential are solely due to external factors like a surrounding dielectric.
  • The formula for the conductor's energy is \( W = \frac{1}{2} QV \).
Once placed in a dielectric, the potential \( V \) and therefore the energy \( W \) decreases.
Potential in Dielectrics
The potential within a dielectric is notably affected by the dielectric constant. When a charged object is immersed in a dielectric, the potential is reduced by the factor of its dielectric constant \( K \).
The original potential \( V_{0} \) becomes \( \frac{V_{0}}{K} \) once inside a dielectric material.
  • This reduction is because the dielectric diminishes the external electric field, thereby reducing potential.
  • The formula shows this effect: \( V = \frac{V_{0}}{K} \).
Consequently, the energy stored within a dielectric is altered based on its potential decrease, as seen in the reduced electrostatic energy equation \( W_d = \frac{W_{0}}{K} \).

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Most popular questions from this chapter

A condenser of capacity \(50 \mu \mathrm{F}\) is charged to \(10 \mathrm{~V}\). Its energy is equal to: (a) \(2.5 \times 10^{-3} \mathrm{~J}\) (b) \(2.5 \times 10^{-4} \mathrm{~J}\) (c) \(5 \times 10^{-2} \mathrm{~J}\) (d) \(1.25 \times 10^{-8} \mathrm{~J}\)

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

A potential difference of \(500 \mathrm{~V}\) is applied to a parallel plate condenser. The separation between plates is \(2 \times 10^{-3} \mathrm{~m}\). The plates of the condenser are vertical. An electron is projected vertically upwards between the plates with a velocity of \(10^{5} \mathrm{~m} / \mathrm{s}\) and it moves undeflected between the plates. The magnetic field acting perpendicular to the electric field has a magnitude of (a) \(1.5 \mathrm{~Wb} / \mathrm{m}^{2}\) (b) \(2.0 \mathrm{~Wb} / \mathrm{m}^{2}\) (c) \(2.5 \mathrm{~Wb} / \mathrm{m}^{2}\) (d) \(3.0 \mathrm{~Wb} / \mathrm{m}^{2}\)

A capacitor is charged by using a battery, which is then disconnected. A dielectric slab is then slided between the plates which results in: (a) reduction of charge on the plates and increase of potential difference across the plates (b) increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates (c) decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates (d) none of the above

A capacitor of capacitance \(C\) is charged to a potential difference \(V_{0}\). The charged battery is disconnected and the capacitor is connected to a capacitor of unknown capacitance \(C_{x}\). The potential difference across the combination is \(V\). The value of \(C_{r}\) should be : (a) \(\frac{C\left(V_{0}-V\right)}{V}\) (b) \(\frac{C\left(V-V_{0}\right)}{V}\) (c) \(\frac{\mathrm{CV}}{V_{0}}\) (d) \(\frac{C V_{0}}{V}\)
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