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What is the capacitance of a large Van de Graaff generator's terminal, given that it stores \(8.00 \mathrm{mC}\) of charge at a voltage of \(12.0 \mathrm{MV}\) ?

Short Answer

Expert verified
The capacitance of the Van de Graaff generator's terminal is approximately \( \frac{2}{3} \times 10^{-9}\) Farads.

Step by step solution

01

Convert units

Firstly, let's convert the given units to SI units: Charge (Q): 8.0 mC = 8.0 * 10^(-3) C (Since 1 mC = 10^(-3) C) Voltage (V): 12.0 MV = 12.0 * 10^6 V (Since 1 MV = 10^6 V)
02

Plug the values into the capacitance formula

Now, we can use the capacitance formula: C = Q / V Substitute Q and V with their converted values: C = (8.0 * 10^(-3) C) / (12.0 * 10^6 V)
03

Simplify the expression and find the capacitance

Now, let's simplify the expression: C = (8.0 * 10^(-3)) / (12.0 * 10^6) = (2/3) * 10^(-9) F (We simplified by dividing the numbers and simplifying the powers of 10) So, the capacitance of the Van de Graaff generator's terminal is approximately \( \frac{2}{3} \times 10^{-9}\) Farads.

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Key Concepts

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

Van de Graaff generator
A Van de Graaff generator is a fascinating device that is often used to demonstrate high voltage physics. It has a recognizable design with a large hollow metallic dome mounted on top of a column. Inside this column, a moving belt transports charge to the dome, allowing it to accumulate high voltage. This accumulation of charge is what gives the Van de Graaff generator its primary function of storing charge and achieving high voltages.

The Van de Graaff generator is typically used in educational settings to provide a safe way to explore electrostatics - the study of stationary electric charges. Due to its ability to develop very high voltages without a corresponding high energy, it is widely valued in both academic demonstrations and playful scientific experiments. Its design makes use of principles like charging by induction and insulation, which help to maintain the charge until it is purposefully discharged.
SI units conversion
Converting units to the International System of Units (SI) is crucial in physics as it allows for standardization and comparison. In the original exercise, we deal with charge measured in milliCoulombs (mC) and voltage in MegaVolts (MV). To use these values in calculations, we need to convert them to Coulombs (C) and Volts (V), respectively.

When converting, remember:
  • 1 milliCoulomb (mC) is equal to \(10^{-3}\) Coulombs (C).
  • 1 MegaVolt (MV) is equal to \(10^6\) Volts (V).
By using these conversion factors, measuring charge becomes \(8.0 \times 10^{-3}\) C and voltage becomes \(12.0 \times 10^6\) V, as shown in the solution. Converting to SI units ensures precision and consistency in engineering and scientific calculations.
Capacitance formula
The capacitance formula is an essential relationship in physics and electrical engineering. It relates the charge stored in a system to the voltage across it. Capacitance (\(C\)) is defined as the ratio of the quantity of charge (\(Q\)) stored on a capacitor to the voltage (\(V\)) applied across it. Mathematically, this is expressed as \[ C = \frac{Q}{V} \].

Using this formula, you can determine the capacitance of devices like capacitors or terminals of generators - such as a Van de Graaff generator. This equation shows that for a given charge, a higher voltage decreases capacitance, while a lower voltage increases it. Understanding the capacitance formula helps in designing circuits and electrical systems according to desired parameters, ensuring functionality and safety.
Charge storage
Charge storage is a fundamental concept in both physics and electrical engineering. Capacitors and other charge-storing devices, such as the Van de Graaff generator, are designed to store electric charge. The amount of charge they can store depends on their capacitance and the voltage across them.

Devices that store charge can then release it as needed. This principle is utilized in smoothing out power supplies, filtering signals, and powering applications temporarily without continuous supply. In the context of the Van de Graaff generator exercise, the stored charge is 8.00 mC (milliCoulombs), illustrating its ability to hold and maintain a charge until discharged. This ability makes it a useful tool for studying electrical properties and behaviors in a very tangible way.
Voltage conversion
Voltage conversion refers to changing from one unit of voltage measurement to another. It's often necessary when working with devices and systems that use different scales for measuring voltage. In the SI system, the basic unit of voltage is the Volt (V), but often you'll encounter kilovolts (kV) and megavolts (MV), especially in high voltage applications.

The process is simple:
  • To convert from MegaVolts to Volts, multiply the number of MegaVolts by \(10^6\). For example, 12.0 MV becomes 12.0 \times 10^6 V.
  • This ensures that all values are converted to a standard format, easing calculations and improving clarity.
Proper voltage conversion is essential to ensure that calculations in physics and engineering are precise, resulting in safe and efficient systems.

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Most popular questions from this chapter

Three capacitors having capacitances 8.4, 8.4, and \(4.2 \mu \mathrm{F}\) are connected in series across a \(36.0-\mathrm{V}\) potential difference. (a) What is the total energy stored in all three capacitors? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged plates connected together. What is the total energy now stored in the capacitors?

If you wish to store a large amount of charge in a capacitor bank, would you connect capacitors in series or in parallel? Explain.

Some cell walls in the human body have a layer of negative charge on the inside surface. Suppose that the surface charge densities are \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is \(5.0 \times 10^{-9} \mathrm{m}\) thick, and the cell wall material has a dielectric constant of \(\kappa=5.4 .\) (a) Find the magnitude of the electric field in the wall between two charge layers. (b) Find the potential difference between the inside and the outside of the cell. Which is at higher potential? (c) A typical cell in the human body has volume \(10^{-16} \mathrm{m}^{3}\) Estimate the total electrical field energy stored in the wall of a cell of this size when assuming that the cell is spherical. (Hint: Calculate the volume of the cell wall.)

Two flat plates containing equal and opposite charges are separated by material 4.0 mm thick with a dielectric constant of \(5.0 .\) If the electrical field in the dielectric is \(1.5 \mathrm{MV} / \mathrm{m},\) what are (a) the charge density on the capacitor plates, and (b) the induced charge density on the surfaces of the dielectric?

Find the charge stored when \(5.50 \mathrm{V}\) is applied to an 8.00-pF capacitor.

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