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An empty capacitor has a capacitance of \(3.2 \mu \mathrm{F}\) and is connected to a \(12-\mathrm{V}\) battery. A dielectric material \((\kappa=4.5)\) is inserted between the plates of this capacitor. What is the magnitude of the surface charge on the dielectric that is adjacent to either plate of the capacitor?

Short Answer

Expert verified
The surface charge on the dielectric is \(1.34 \times 10^{-4} C\).

Step by step solution

01

Calculate Capacitance with Dielectric

First, calculate the new capacitance when the dielectric is in place. The formula to find the new capacitance \( C' \) is:\[ C' = \kappa \cdot C \]where \( \kappa = 4.5 \) and \( C = 3.2 \mu F \). Therefore, \[ C' = 4.5 \times 3.2 \mu F = 14.4 \mu F. \]
02

Determine the Charge on the Capacitor

Next, use the capacitance formula \( Q = C \cdot V \) to find the charge \( Q \) on the capacitor with the dielectric. Substitute \( C' = 14.4 \mu F \) and \( V = 12 V \) into the formula:\[ Q = 14.4 \times 10^{-6} F \times 12 V = 1.728 \times 10^{-4} C. \]
03

Find the Surface Charge on the Dielectric

The surface charge \( Q_d \) on the dielectric can be found using the relation:\[ Q_d = Q \left(1 - \frac{1}{\kappa}\right) \]Substitute \( Q = 1.728 \times 10^{-4} C \) and \( \kappa = 4.5 \) into the formula:\[ Q_d = 1.728 \times 10^{-4} C \left(1 - \frac{1}{4.5}\right) = 1.728 \times 10^{-4} C \times 0.7778 = 1.34 \times 10^{-4} C. \]

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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, denoted by the Greek letter \( \kappa \), plays an essential role in capacitors. It represents a measure of a material's ability to reduce the electric field inside a capacitor. When a dielectric is introduced between the plates of a capacitor, it enhances the capacitor's capacitance. This happens because the dielectric material polarizes in response to the electric field, thereby reducing the field and allowing the capacitor to store more charge for a given voltage.
This increase in the ability to store charge can be quantified by multiplying the initial capacitance (without the dielectric) by the dielectric constant. Thus, the equation is \( C' = \kappa \cdot C \), where \( C' \) is the new capacitance with the dielectric, \( C \) is the original capacitance, and \( \kappa \) is the dielectric constant.
Surface charge
The concept of surface charge on a dielectric is quite fascinating. It refers to the distribution of charge that occurs on the surface of the dielectric when it is placed between the capacitor's plates. This surface charge arises due to the polarization of the dielectric material.
When the dielectric is introduced into the electric field created by the capacitor's plates, its molecules align in such a way that positive charges face one plate and negative charges face the other. This alignment creates an internal electric field within the dielectric, counteracting part of the overall field between the plates, which in turn increases capacitance.
The surface charge \( Q_d \) can be calculated using the formula:
  • \( Q_d = Q \left(1 - \frac{1}{\kappa}\right) \)
where \( Q \) is the charge on the plates and \( \kappa \) is the dielectric constant. This formula tells us how much charge is effectively neutralized by the dielectric material's polarization.
Capacitance formula
The capacitance formula describes how much charge a capacitor can store per volt of potential difference across its plates. It is a foundational principle in understanding capacitors. The basic formula used to determine the charge \( Q \) stored in a capacitor is \( Q = C \cdot V \), where:
  • \( Q \) is the charge in coulombs.
  • \( C \) is the capacitance in farads.
  • \( V \) is the voltage in volts.
The formula connects the physical ability of the capacitor to store charge with the voltage applied across it.
In the case of having a dielectric, the capacitance increases by the factor of the dielectric constant \( \kappa \), so the updated formula becomes \( Q = C' \cdot V \), where \( C' = \kappa \cdot C \). This shows that the presence of a dielectric can significantly increase the amount of charge stored for the same voltage.

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Most popular questions from this chapter

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of \(0.75 \mathrm{mm} .\) When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{V} / \mathrm{m} .\) What is the magnitude of the potential difference \(\Delta V\) between the conductors?

During a particular thunderstorm, the electric potential difference between a cloud and the ground is \(V_{\text {cloud }}-V_{\text {ground }}=1.3 \times 10^{8} \mathrm{V},\) with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?

Two capacitors are identical, except that one is empty and the other is filled with a dielectric \((\kappa=4.50) .\) The empty capacitor is connected to a \(12.0-\mathrm{V}\) battery. What must be the potential difference across the plates of the capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacitor?

A particle with a charge of \(-1.5 \mu \mathrm{C}\) and a mass of \(2.5 \times 10^{-6} \mathrm{kg}\) is released from rest at point \(A\) and accelerates toward point \(B\), arriving there with a speed of \(42 \mathrm{m} / \mathrm{s}\). The only force acting on the particle is the electric force. (a) Which point is at the higher potential? Give your reasoning. (b) What is the potential difference \(V_{\mathrm{B}}-V_{\mathrm{A}}\) between \(\mathrm{A}\) and \(\mathrm{B} ?\)

Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length \(0.480 \mathrm{m}\). A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right)\), with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

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