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Two conductors having net charges of \(+10.0 \mu \mathrm{C}\) and \(-10.0 \mu \mathrm{C}\) have a potential difference of \(10.0 \mathrm{V}\) between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to \(+100 \mu \mathrm{C}\) and \(-100 \mu \mathrm{C} ?\)

Short Answer

Expert verified
The capacitance of the system is \(1 \mu F\) and the potential difference between the two conductors when the charges on each are increased to \(+100 \mu C\) and \(-100 \mu C\) is \(100 V\).

Step by step solution

01

Understanding the formulas

First, understand that the concept of capacitance (\(C\)) is defined by the relationship \(Q = VC\), where \(Q\) is the charge stored in one of the conductors and \(V\) is the potential difference between the conductors. Solving this equation for \(C\) gives us \(C = Q/V\).
02

Calculating the capacitance

Plug the given values for the charge (\(Q = 10.0 \mu C\)) and potential difference (\(V = 10.0 V\)) into the equation for \(C\). This gives \(C = (10.0 \mu C) / (10.0 V) = 1 \mu F\). Thus, the capacitance of the system is \(1 \mu F\).
03

Finding the new potential difference

Next, use the same relationship \(Q = VC\) to find the new potential difference when the charge is increased to \(100 \mu C\). Doing so gives \(V = Q/C = (100 \mu C) / (1 \mu F) = 100 V\). So the potential difference between the two conductors when the charges on each are increased to \(+100 \mu C\) and \(-100 \mu C\) is \(100 V\).

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Electrons have a negative charge, while protons have a positive charge. These charges are central to many electric phenomena, such as how electric currents flow through wires.

Here are a few key points about electric charge:
  • Charges exist in discrete quantities, often measured in Coulombs (\(C\)).
  • There are two types of charges, positive and negative, which can attract or repel each other depending on their nature.
  • Charge conservation states that the total charge in an isolated system remains constant regardless of any changes within the system.
In the context of capacitors, electric charge is essential because it is stored between a pair of conductors. In the original exercise, the conductors have charges of \(+10.0 \mu C\) and \(-10.0 \mu C\), meaning they are equal in magnitude but opposite in sign. This kind of setup is typical for capacitors, as it allows them to store energy efficiently.
Potential Difference
The potential difference, often referred to as voltage, is the measure of the work needed to move a unit charge from one point to another in an electric field. It is the driving force that allows the movement of electric charge, similar to how water flows from a high point to a low point due to gravity.

Important aspects of potential difference include:
  • It is measured in Volts (\(V\)), where 1 Volt equals 1 Joule per Coulomb.
  • It determines the amount of energy transferred per unit charge as it moves through the circuit.
  • In electrical systems, increasing the potential difference results in a stronger electric field, which can move charges more efficiently.
In the exercise, the potential difference between the conductors was initially \(10.0 V\). This difference increases to \(100 V\) when the net charges on the conductors are increased to \(+100 \mu C\) and \(-100 \mu C\). This demonstrates how potential difference relates directly to the amount of charge and capacitance in a system.
Capacitor
A capacitor is a device that stores energy in the electric field between a pair of conductors commonly known as "plates." The stored energy can then be discharged to provide a stabilizing function in circuits, pulsed power, or filtering.

Some essential features of capacitors include:
  • They consist of two conductive plates separated by an insulating material known as a dielectric.
  • The capacitance (\(C\)) of a capacitor is determined by its ability to store electric charge. It is calculated as \(C = Q/V\), where \(Q\) is the charge, and \(V\) is the potential difference.
  • Capacitors are measured in Farads (\(F\)), but practical capacitors are often in microfarads (\(\mu F\)), nanofarads (\(nF\)), or picofarads (\(pF\)).
In the specific problem from the exercise, we calculated the system's capacitance to be \(1 \mu F\). This value indicates the relationship between the electric charge and the potential difference across the conductors, offering a way to understand how energy storage and discharge operate within the circuit.

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Most popular questions from this chapter

Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of \(1.75 \mathrm{cm}^{2}\) and plate separation of \(0.0400 \mathrm{mm}\).

A detector of radiation called a Geiger tube consists of a closed, hollow, conducting cylinder with a fine wire along its axis. Suppose that the internal diameter of the cylinder is \(2.50 \mathrm{cm}\) and that the wire along the axis has a diameter of \(0.200 \mathrm{mm} .\) The dielectric strength of the gas between the central wire and the cylinder is \(1.20 \times 10^{6} \mathrm{V} / \mathrm{m} .\) Calculate the maximum potential difference that can be applied between the wire and the cylinder before breakdown occurs in the gas.

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) \(\mathrm{A}\) sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform clectric ficld at nearby points, with magnitude \(E=Q / 2 A \epsilon\). (b) Two large sheets of area \(A\), carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Qd/Ae. (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d\).

A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group?

A parallel-plate capacitor has a charge \(Q\) and plates of area \(A\). What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is \(E=Q / A \epsilon_{0},\) you might think that the force is \(F=Q E=Q^{2} / A \epsilon_{0} .\) This is wrong, because the field \(E\) includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually \(F=Q^{2} / 2 \epsilon_{0} A .\) (Suggestion: Let \(C=\epsilon_{0} A / x\) for an arbitrary plate separation \(x ;\) then require that the work done in separating the two charged plates be \(W=\int F \, d x\).) The force exerted by one charged plate on another is sometimes used in a machine shop to hold a workpiece stationary.

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