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Chapter 25: Problem 20

What is the charge on a 300 -pF capacitor when it is charged to a voltage of \(1.0 \mathrm{kV} ?\) \(q=C V=\left(300 \times 10^{-12} \mathrm{~F}\right)(1000 \mathrm{~V})=3.0 \times 10^{-7} \mathrm{C}=0.30 \mu \mathrm{C}\)

Short Answer

Expert verified
The charge on the capacitor is \(0.30\, \mu\text{C}\).

Step by step solution

01

Understand the Formula

The formula to find the charge \(q\) on a capacitor is given by \(q = C \times V\), where \(C\) is the capacitance in farads and \(V\) is the voltage in volts.
02

Convert Capacitance and Voltage

We have a capacitance \(C = 300\, \text{pF}\) (picoFarads) which is equivalent to \(300 \times 10^{-12}\, \text{F}\), and a voltage \(V = 1.0\, \text{kV}\) (kiloVolts) which is equivalent to \(1000\, \text{V}\).
03

Plug the Values into the Formula

Using the formula \(q = C \times V\), plug in the values: \(q = (300 \times 10^{-12}\, \text{F})(1000\, \text{V})\).
04

Calculate the Charge

Calculate the result: \(q = 300 \times 10^{-12} \times 1000 = 3.0 \times 10^{-7}\, \text{C}\).
05

Convert to Microcoulombs

Convert the charge into microcoulombs: \(3.0 \times 10^{-7}\, \text{C} = 0.30 \, \mu \text{C}\) because \(1 \mu \text{C} = 10^{-6} \text{C}\).

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Key Concepts

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

Capacitance Conversion
Understanding how to convert capacitance measurements is crucial when working with capacitors. Capacitance, denoted by the letter \(C\), is usually measured in farads (\(F\)). However, typical capacitance values for most practical capacitors are often in smaller units such as nanofarads (\(nF\)), microfarads (\(\mu F\)), or picofarads (\(pF\)).
To convert from picofarads (\(pF\)) to farads (\(F\)), it's important to remember the relationship:
  • \(1 \, pF = 1 \, \times 10^{-12} \, F\)
For example, a capacitor with a capacitance of 300 picofarads is converted to farads by multiplying the capacitance by \(10^{-12}\). Therefore, \(300 \, pF = 300 \, \times 10^{-12} \, F\). By performing this conversion, you can fit the values directly into equations that calculate electrical properties such as charge.
Voltage Conversion
Voltage is another critical part of calculating the charge on a capacitor. It's often necessary to convert voltage values for use in equations. Voltage (\(V\)) is the electric potential difference and is typically measured in volts (\(V\)).
Sometimes, voltage is given in kilovolts (\(kV\)), which requires you to convert it into volts. The conversion is straightforward:
  • \(1 \, kV = 1000 \, V\)
For instance, if a capacitor is charged to a voltage of \(1.0 \, kV\), you'll convert it to volts: \(1.0 \, kV = 1000 \, V\). Converting the voltage into volts ensures that all units are consistent when using the charge formula.
Charge Formula
The fundamental relationship used to determine the charge stored in a capacitor is given by the formula \(q = C \times V\). In this formula, \(q\) represents the charge in coulombs, \(C\) is the capacitance in farads, and \(V\) is the voltage in volts.
Here's how it works:
  • Multiply the capacitance (in farads) by the voltage (in volts). This gives you the charge in coulombs.
For example, with a capacitance of \(300 \, \times 10^{-12} \, F\) and a voltage of \(1000 \, V\), the charge can be calculated as \(q = (300 \, \times 10^{-12} \, F)(1000 \, V) = 3.0 \, \times 10^{-7} \, C\). This formula helps in determining the amount of charge a capacitor holds for a given voltage.
Microcoulomb Conversion
After calculating the charge in coulombs, it is often useful to convert it to microcoulombs for easier interpretation, especially when dealing with small charges. The microcoulomb (\(\mu C\)) is a subunit of the coulomb, with the conversion relation as follows:
  • \(1 \, \mu C = 10^{-6} \, C\)
To convert from coulombs to microcoulombs, multiply by \(10^{6}\). For instance, a charge of \(3.0 \, \times 10^{-7} \, C\) is equivalent to \(0.30 \, \mu C\) since \(3.0 \, \times 10^{-7} \, C = 0.30 \, \times 10^{-6} \, C\).
This conversion is especially handy in fields like electronics, where small charge units are more practical.

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Most popular questions from this chapter

A potential difference of \(24 \mathrm{kV}\) maintains a downward-directed electric field between two horizontal parallel plates separated by \(1.8 \mathrm{~cm}\) in vacuum. Find the charge on an oil droplet of mass \(2.2 \times\) \(10^{-13}\) kg that remains stationary in the field between the plates.

Two capacitors \((0.30\) and \(0.50 \mu \mathrm{F}\) ) are connected in parallel. \((a)\) What is their equivalent capacitance? A charge of \(200 \mu \mathrm{C}\) is now placed on the parallel combination. (b) What is the potential difference across it? (c) What are the charges on the capacitors?

An electron starts from rest and falls through a potential rise of 80 V. What is its final speed? Positive charges fall through potential drops; negative charges, such as electrons, fall through potential rises. Change in \(\mathrm{PE}_{E}=V q=(80 \mathrm{~V})\left(-1.6 \times 10^{-19} \mathrm{C}\right)=-1.28 \times 10^{-17} \mathrm{~J}\) This lost \(\mathrm{PE}_{E}\) appears as KE of the electron:= and $$ \begin{array}{c} \mathrm{PE}_{\varepsilon} \text { lost }=\text { KE gained } \\\ 1.28 \times 10^{-17} \mathrm{~J}=\frac{1}{2} m v_{f}^{2} \frac{1}{2} m v_{i}^{2}=\frac{1}{2} m v_{f}^{2}-0 \\ v_{f}=\sqrt{\frac{\left(1.28 \times 10^{-17} \mathrm{~J}(2)\right.}{9.1 \times 10^{-31} \mathrm{~kg}}}=5.3 \times 10^{6} \mathrm{~m} / \mathrm{s} \end{array} $$

The nucleus of a tin atom in vacuum has a charge of \(+50 e .(a)\) Find the absolute potential \(V\) at a radial distance of \(1.0 \times 10^{-12} \mathrm{~m}\) from the nucleus. ( \(b\) ) If a proton is released from this point, how fast will it be moving when it is \(1.0 \mathrm{~m}\) from the nucleus? (a) \(V=k_{0} \frac{q}{f}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2}\right) \frac{(50)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{10^{-12} \mathrm{~m}}=72 \mathrm{kV}\) (b) The proton is repelled by the nucleus and flies out to infinity. The absolute potential at a point is the potential difference between the point in question and infinity. Hence, there is a potential drop of \(72 \mathrm{kV}\) as the proton flies to infinity. Usually we would simply assume that \(1.0 \mathrm{~m}\) is far enough from the nucleus to consider it to be at infinity. But, as a check, compute \(V\) at \(r=1.0 \mathrm{~m}\) : $$ V_{1 \mathrm{~m}}=k_{0} \frac{q}{r}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{(50)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{1.0 \mathrm{~m}}=7.2 \times 10^{-8} \mathrm{~V} $$ which is essentially zero in comparison with \(72 \mathrm{kV}\). As the proton falls through \(72 \mathrm{kV}\), $$ \begin{aligned} \text { KE gained } &=\mathrm{PE}_{E} \text { lost } \\ \frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} &=q V \\ \frac{1}{2}\left(1.67 \times 10^{-27} \mathrm{~kg}\right) v_{f}^{2}-0 &=\left(1.6 \times 10^{-19} \mathrm{C}\right)(72000 \mathrm{~V}) \end{aligned} $$ from which \(v_{f}=3.7 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Three capacitors \((2.00 \mu \mathrm{F}, 5.00 \mu \mathrm{F}\), and \(7.00 \mu \mathrm{F})\) are connected in parallel. What is their equivalent capacitance?
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