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Chapter 25: Problem 92

A parallel plate condenser is charged by a battery. The battery is removed and a thick.glass slab is placed between the plates. Now: (a) the capacity of the condenser is increased (b) the potential across the plates is decreased (c) the electric field between the plates is decreased (d) all of the above

Short Answer

Expert verified
(d) all of the above

Step by step solution

01

Understanding the Problem

The problem involves a parallel plate capacitor that has been charged and then has a thick glass slab placed between its plates after disconnecting it from the battery. We need to analyze the effects of inserting the glass slab.
02

Concept of Capacitance with Dielectrics

A glass slab is a dielectric material. When a dielectric is inserted between the plates of a capacitor, the capacitance increases. This is because the dielectric effectively reduces the electric field within the capacitor, allowing it to store more charge for a given potential difference.
03

Analyzing the Potential Across the Plates

The battery has been removed, so the charge on the plates remains constant. The formula for capacitance with a dielectric is given by \( C' = \kappa C \), where \( \kappa \) is the dielectric constant and \( C \) is the initial capacitance. The potential difference \( V' \) is related to the original potential \( V \) by \( C'V' = CV \). Since \( C' > C \), this means \( V' < V \). Thus, the potential across the plates decreases.
04

Examining the Electric Field

The electric field \( E \) is related to the potential difference \( V \) by \( E = \frac{V}{d} \), where \( d \) is the separation between the plates. Since the potential decreases upon inserting the dielectric, the electric field also decreases.
05

Drawing Conclusions

The insertion of a glass slab increases the capacitance, decreases the potential across the plates, and, consequently, decreases the electric field. Therefore, all the given statements (a, b, c) are true.

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Key Concepts

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

Capacitance Increase
When you insert a dielectric material like a thick glass slab between the plates of a capacitor, something interesting happens: the capacitance of the capacitor increases.
Capacitance is essentially the capacitor's ability to store an electric charge. Typically, with parallel plate capacitors, capacitance without any dielectric material is given by the formula:
  • \( C = \frac{\varepsilon_0 A}{d} \)
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the separation between the plates.
Incorporating Dielectrics
  • When a dielectric is added, the capacitance formula modifies to \( C' = \kappa C \), where \( \kappa \) is the dielectric constant specific to the material used鈥攍ike glass.
  • This means the capacitance increases by the factor \( \kappa \), allowing the capacitor to store more charge at the same voltage.
This increase in capacitance is a crucial factor in designing circuits, as it means more charge can be stored without necessarily increasing the size of the capacitor.
Electric Potential Decrease
Once you place the dielectric glass slab between the capacitor plates and have already disconnected the battery, you're left with a constant charge on the plates.
But what happens to the electric potential or voltage across the plates? It decreases.
Why Does This Happen?
  • With the increase in capacitance \( C' = \kappa C \), the relationship between charge \( Q \), capacitance \( C \), and voltage \( V \) is given by \( Q = C' V' \).
  • Since the charge \( Q \) remains constant and \( C' \) is greater than \( C \), the potential difference \( V' \) has to decrease to maintain \( Q = CV \).
This lowering of voltage is a direct consequence of the capacitor's ability to store more energy thanks to the dielectric. This is especially useful in applications where high energy storage with lower voltages is desired.
Electric Field Reduction
The electric field (\( E \)) between the plate's changes as a result of inserting a dielectric. Originally, without the glass, the electric field can be thought of as the movement of charge due to voltage, expressed as \( E = \frac{V}{d} \).
But what happens when you add a dielectric and how does that affect the electric field?
Effects of Dielectric Materials
  • The dielectric reduces the voltage \( V \), which in turn reduces the electric field \( E \).
  • Inserting the glass slab results in the field strength decreasing because \( E' = \frac{V'}{d} \) and since \( V' < V \), it's clear that \( E' < E \).
This reduction in electric field strength is advantageous in preventing potential breakdowns in the material by lowering the risk of sparking or discharges, thus enhancing the capacitor's resilience.

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Most popular questions from this chapter

Force acting upon a charged particle, kept between the plates of a charged condenser is \(F\). If one of the plates of the condenser is removed, force acting on the same particle will become: (a) 0 (b) \(F / 2\) (c) \(F\) (d) \(2 F\)

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

Three identical capacitors are connected together differently. For the same voltage to every combination, the one that stores maximum energy is: (a) the three capacitors in series (b) the three capacitors in parallel (c) two capacitors in series with third capacitor in parallel with it (d) two capacitors in parallel with the third capacitors in series with it

Two conducting plates \(x\) and \(y\), each having large surface area \(A\) (on the side) are placed parallel to each other. The plate \(x\) is given a charge \(Q\) whereas the other is neutral. The electric field at a point in between the plates is given by: (a) \(\frac{Q}{2 A}\) (b) \(\frac{Q}{2 A \varepsilon_{0}}\) towards left (c) \(\frac{Q}{2 A \varepsilon_{0}}\) towards right (d) \(\frac{Q}{2 \varepsilon_{0}}\) towards right

When a capacitor is connected to a battery, then: (a) no current flows in the circuit at all (b) the current flows in the circuit for some time then decreases to zero (c) the current keeps on increasing, reaching a maximum value when the capacitor is charged to the voltage of the battery (d) an alternating current flows in the circuit
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