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Chapter 30: Problem 21

Two electrodes connected to a \(9.0 \mathrm{V}\) battery are charged to 卤45 nC. What is the capacitance of the electrodes?

Short Answer

Expert verified
After performing the calculation for \(C\), you get approximately \(5 脳 10^{-9} F\).

Step by step solution

01

Understand what the question asks for

This question asks for the capacitance value of the two electrodes. The known values from the problem description are: charge (\(Q\)) stored which is 卤45 nC and a potential difference (\(V\)) of 9.0 V. We know that capacitance (\(C\)) can be computed using the formula \(C = Q/V\).
02

Convert the charge from nC to C

In order for your units to be comparable and accurate, you should convert the charge given in \(nC\) (nanocoulombs) to \(C\) (coulombs). 1 \(nC\) is equal to \(10^{-9} C\). Therefore, \(45 nC = 45 脳 10^{-9} C\).
03

Substitute the present values into the formula

Using the formula \(C = Q/V\), substitute \(Q = 45 脳 10^{-9}C\) and \(V = 9.0 V\). You can then compute the capacitance as \(C = (45 脳 10^{-9} C) / 9.0 V\).

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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. It refers to the amount of electric energy stored in an object. There are two types of electric charges: positive and negative. They attract each other, while like charges repel. This behavior is a key principle of electrostatics.
  • Electric charges are measured in coulombs (C).
  • Every charged particle carries a basic unit of charge, either positive or negative.
  • In the problem, each electrode is charged to 卤45 nC, where 'n' indicates nano, or one-billionth of a coulomb.
  • To convert nanocoulombs to coulombs, remember that 1 nC = 10^-9 C, a small charge significant in electrostatics.
Understanding electric charge helps us calculate many electric phenomena, such as finding the capacitance in this exercise.
Potential Difference
Potential difference, often referred to as voltage, is the measure of electrical potential energy between two points in an electric field. It's like the force that pushes electrical charges to move through a circuit. The greater the potential difference, the stronger the "push."
  • Measured in volts (V).
  • In this problem, the given potential difference is 9.0 V.
  • A higher potential difference generally means a stronger electric field and more energy provided to the charges.
  • Potential difference is essential in calculating the capacitance, acting as the medium allowing a charge to store energy.
The concept is critical in circuits, where it determines how much electrical energy is supplied and consumed.
Electrostatics
Electrostatics is a branch of physics that studies stationary or slow-moving electric charges. It explores how and why charges interact, focusing on forces and fields involved when charges are held still. In this context, capacitance is a vital concept.
  • Electrostatics includes the study of forces exerted by electric charges.
  • Capacitance is the ability of a system to store an electric charge.
  • The formula to find capacitance is \(C = \frac{Q}{V}\), where \(C\) is capacitance, \(Q\) is charge, and \(V\) is potential difference.
  • This formula allows us to calculate how well a capacitance can store an electric charge when connected to an electric potential difference.
By understanding electrostatics, students can understand many phenomena, such as how capacitors work in electrical devices.

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Most popular questions from this chapter

What is the emf of a battery that will charge a \(2.0 \mu\) F capacitor to \(\pm 48 \mu \mathrm{C} ?\)

The radiation detector known as a Geiger counter uses a closed, hollow, cylindrical tube with an insulated wire along its axis. Suppose a Geiger tube, as it's called, has a 1.0 -mm-diameter wire in a tube with a 25 mm inner diameter. The tube is filled with a low-pressure gas whose dielectric strength is \(1.0 \times\) \(10^{6} \mathrm{V} / \mathrm{m} .\) What is the maximum potential difference between the wire and the tube?

Two \(5.0 \mathrm{mm} \times 5.0 \mathrm{mm}\) electrodes with a 0.10 -mm-thick sheet of Mylar between them are attached to a \(9.0 \mathrm{V}\) battery. Without disconnecting the battery, the Mylar is withdrawn. (Very small spacers keep the electrode separation unchanged.) What are the charge, potential difference, and electric field (a) before and (b) after the Mylar is withdrawn?

The electric potential along the \(x\) -axis is \(V=100 x^{2}\) V, where \(x\) is in meters. What is \(E_{x}\) at (a) \(x=0\) m and (b) \(x=1\) m?

The electric potential in a region of space is \(V=100\left(x^{2}-\right.\) \(y^{2}\) ) \(V,\) where \(x\) and \(y\) are in meters. a. Draw a contour map of the potential, showing and labeling the \(-400 \mathrm{V},-100 \mathrm{V}, 0 \mathrm{V},+100 \mathrm{V},\) and \(+400 \mathrm{V}\) equipotential surfaces. b. Find an expression for the electric field \(\vec{E}\) at position \((x, y)\) c. Draw the electric field lines on your diagram of part a.
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