/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Determine the charge on each pla... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the charge on each plate of a \(0.050-\mu \mathrm{F}\) parallel-plate capacitor when the potential difference between the plates is \(200 \mathrm{~V}\).

Short Answer

Expert verified
Each plate has a charge of \( 10 \mu C \).

Step by step solution

01

Identify the formula for the charge on a capacitor

The charge \( Q \) on a capacitor is given by the formula \( Q = C \times V \), where \( C \) is the capacitance and \( V \) is the potential difference.
02

Substitute the given values into the formula

Use the values given: \( C = 0.050 \mu F \) and \( V = 200 \mathrm{~V} \). First, convert the capacitance from microfarads to farads: \( 0.050 \mu F = 0.050 \times 10^{-6} F \). Substitute these into the formula: \( Q = (0.050 \times 10^{-6}) F \times 200 \mathrm{~V} \).
03

Perform the calculation

Calculate the charge: \( Q = 0.050 \times 10^{-6} \times 200 = 10 \times 10^{-6} \), which simplifies to \( 10 \mu C \).
04

Interpret the result

The resulting charge on each plate of the capacitor is \( 10 \mu C \). This means each plate has a charge of \( 10 \mu C \), with one plate being positively charged and the other being negatively charged.

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Key Concepts

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

Charge Calculation
Calculating the charge on a capacitor involves understanding how capacitance and potential difference work together to store electric charge. Essentially, a capacitor has two plates that hold equal and opposite charges. The amount of charge a capacitor can store is directly proportional to both its capacitance (the ability to store charge) and the potential difference (voltage) across its plates.

To calculate the charge (\( Q \)) on a capacitor, you use the formula: \( Q = C \times V \).
  • \( C \) is the capacitance of the capacitor. Capacitance quantifies a capacitor's ability to store a charge per unit voltage.
  • \( V \) is the potential difference, or voltage, across the capacitor's plates.
  • Use the formula to find the charge by substituting the known values of capacitance and voltage into it.
In our example, we first convert the capacitance from microfarads (\( \mu F \)) to farads by multiplying by \( 10^{-6} \). Then, multiplying the converted capacitance with the voltage gives the charge stored in the capacitor.
Potential Difference
Potential difference, more commonly referred to as voltage, is crucial when working with capacitors. It is the "push" that motivates charge to move across the capacitor plates. This push can be thought of as the energy required to move a charge from one plate to another, and it is measured in volts (\( V \)).

In practical terms, the potential difference determines the amount of energy stored in a capacitor for a given amount of capacitance. The larger the voltage supplied to a capacitor, the greater the charge it can hold, assuming the capacitance remains constant. As such, understanding potential difference helps in predicting how a capacitor will behave in a circuit.
  • A higher potential difference allows for more charge to be stored.
  • It directly affects and is proportional to the charge accumulated on the capacitor.
In the given example, a potential difference of 200 volts across the capacitor plates enables the storing of a charge calculated by multiplying this voltage by the capacitance.
Capacitance
Capacitance is a property of a capacitor that defines its ability to hold or "store" electric charge. The standard unit for capacitance is the farad (\( F \)), however, values are often expressed in microfarads (\( \mu F \)) due to practicality in everyday electronics.

Understanding capacitance involves knowing how design and physical properties influence a capacitor's performance. Factors that affect capacitance include:
  • The surface area of the plates: Larger areas can hold more charge.
  • The distance between plates: Closer plates result in higher capacitance.
  • The dielectric material between the plates: This material impacts how much charge can be stored.
In circuits, a higher capacitance signifies a greater ability to store charge at a given voltage. Thus, when calculating or interpreting results involving capacitors, knowing the capacitance provides insight into how the capacitor will act when a potential difference is applied, as demonstrated in the example exercise.

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Most popular questions from this chapter

A charge of \(0.20 \mu \mathrm{C}\) is \(30 \mathrm{~cm}\) from a point charge of \(3.0 \mu \mathrm{C}\) in vacuum. What work is required to bring the \(0.20-\mu \mathrm{C}\) charge \(18 \mathrm{~cm}\) closer to the \(3.0-\mu \mathrm{C}\) charge?

Four point charges in air are placed at the four corners of a square that is \(30 \mathrm{~cm}\) on each side. Find the potential at the center of the square if \((a)\) the four charges are each \(+2.0 \mu \mathrm{C}\) and \((b)\) two of the four charges are \(+2.0 \mu \mathrm{C}\) and two are \(-2.0 \mu \mathrm{C}\). (a) \(V=k_{0} \sum \frac{q_{i}}{r_{i}}=k_{0} \frac{\sum q_{i}}{r}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{(4)\left(2.0 \times 10^{-6} \mathrm{C}\right)}{(0.30 \mathrm{~m})\left(\cos 45^{\circ}\right)}=3.4 \times 10^{5} \mathrm{~V}\) (b) \(V=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{ \left.(2.0+2.0-2.0-2.0) \times 10^{-6} \mathrm{C}\right)}{(0.30 \mathrm{~m})\left(\cos 45^{\circ}\right)}=0\)

Two capacitors, \(3.0 \mu \mathrm{F}\) and \(4.0 \mu \mathrm{F}\), are individually charged across a 6.0-V battery. After being disconnected from the battery, they are connected together with a negative plate of one attached to the positive plate of the other. What is the final charge on each capacitor? Let \(3.0 \mu \mathrm{F}=\mathrm{C}_{1}\) and \(4.0 \mu \mathrm{F}=\mathrm{C}_{2}\). The situation is shown in Fig. \(25-10\). Before being connected, their charges are $$ \begin{array}{l} q_{1}=C_{1} V=\left(3.0 \times 10^{-6} \mathrm{~F}\right)(6.0 \mathrm{~V})=18 \mu \mathrm{C} \\ q_{2}=\mathrm{C}_{2} V=\left(4.0 \times 10^{-6} \mathrm{~F}\right)(6.0 \mathrm{~V})=24 \mu \mathrm{C} \end{array} $$ These charges partly cancel when the capacitors are connected together. Their final charges are \(q_{1}^{\prime}\) and \(q_{2}^{\prime}\), where $$ q_{1}^{\prime}+q_{2}^{\prime}=q_{2}-q_{1}=6.0 \mu \mathrm{C} $$ Also, the potentials across them are now the same, so that \(V=q / \mathrm{C}\) gives $$ \frac{q_{1}^{\prime}}{3.0 \times 10^{-6} \mathrm{~F}}=\frac{q_{2}^{\prime}}{4.0 \times 10^{-6} \mathrm{~F}} \quad \text { or } \quad q_{1}^{\prime}=0.75 q_{2}^{\prime} $$ Substitution in the previous equation gives $$ 0.75 q_{2}^{\prime}+q_{2}^{\prime}=6.0 \mu \mathrm{C} \quad \text { or } \quad q_{2}^{\prime}=3.4 \mu \mathrm{C} $$ Then \(q_{1}^{\prime}=0.75 q_{2}^{\prime}=2.6 \mu \mathrm{C}\).

A laboratory capacitor consists of two parallel conducting plates, each with area \(200 \mathrm{~cm}^{2}\), separated by a 0.40-cm air gap. ( \(a\) ) Compute its capacitance. ( \(b\) ) If the capacitor is connected across a 500-V source, find the charge on it, the energy stored in it, and the value of \(E\) between the plates. \((c)\) If a liquid with \(K=2.60\) is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500-V source? (a) For a parallel-plate capacitor with air gap, $$ C=K \epsilon_{0} \frac{A}{d}=(1)\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right) \frac{200 \times 10^{-4} \mathrm{~m}^{2}}{4.0 \times 10^{-3} \mathrm{~m}}=4.4 \times 10^{-11} \mathrm{~F}=44 \mathrm{pF} $$ (b) \(q=C V=\left(4.4 \times 10^{-11} \mathrm{~F}\right)(500 \mathrm{~V})=2.2 \times 10^{-8} \mathrm{C}=22 \mathrm{nC}\) $$ \begin{aligned} \text { Energy } &=\frac{1}{2} q V=\frac{1}{2}\left(2.2 \times 10^{-8} \mathrm{C}\right)(500 \mathrm{~V})=5.5 \times 10^{-6} \mathrm{~J}=5.5 \mu \mathrm{J} \\ E &=\frac{V}{d}=\frac{500 \mathrm{~V}}{4.0 \times 10^{-3} \mathrm{~m}}=1.3 \times 10^{5} \mathrm{~V} / \mathrm{m} \end{aligned} $$ (c) The capacitor will now have a capacitance \(K=2.60\) times larger than before. Therefore, $$ q=C V=\left(2.60 \times 4.4 \times 10^{-11} \mathrm{~F}\right)(500 \mathrm{~V})=5.7 \times 10^{-8} \mathrm{C}=57 \mathrm{nC} $$ The capacitor already had a charge of \(22 \mathrm{nC}\), and so \(57 \mathrm{nC}-22 \mathrm{nC}\) or \(35 \mathrm{nC}\) must have been added to it.

Find the electrical potential energy of three point charges placed in vacuum as follows on the \(x\) -axis: \(+2.0 \mu \mathrm{C}\) at \(x=0,+3.0 \mu \mathrm{C}\) at \(x=20 \mathrm{~cm}\), and \(+6.0 \mu \mathrm{C}\) at \(x=50 \mathrm{~cm}\). Take the \(\mathrm{PE}_{E}\) to be zero when the charges are separated far apart. $$ V_{x=0.2}=k_{0} \frac{2.0 \mu \mathrm{C}}{0.20 \mathrm{~m}}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{2 \times 10^{-6} \mathrm{C}}{0.20 \mathrm{~m}}\right)=9.0 \times 10^{4} \mathrm{~V} $$ Therefore the work required to bring in the \(3 \mu \mathrm{C}\) charge is $$ W_{3 \mu \mathrm{C}}=q V_{x=0.2}=\left(3.0 \times 10^{-6} \mathrm{C}\right)\left(9.0 \times 10^{4} \mathrm{~V}\right)=0.270 \mathrm{~J} $$ Finally bring the \(6.0 \mu \mathrm{C}\) charge in to \(x=050 \mathrm{~m}\). The potential there due to the two charges already present is $$ V_{x=0.5}=k_{0}\left(\frac{2.0 \times 10^{-6} \mathrm{C}}{0.50 \mathrm{~m}}+\frac{3.0 \times 10^{-6} \mathrm{C}}{0.30 \mathrm{~m}}\right)=12.6 \times 10^{4} \mathrm{~V} $$ Therefore the work required to bring in the \(6.0 \mu \mathrm{C}\) charge is $$ W_{6 \mu \mathrm{C}}=q V_{x=0.5}=\left(6.0 \times 10^{-6} \mathrm{C}\right)\left(12.6 \times 10^{4} \mathrm{~V}\right)=0.756 \mathrm{~J} $$ Adding the amounts of work required to assemble the charges gives the energy stored in the system: $$ \mathrm{PE}_{E}=0.270 \mathrm{~J}+0.756 \mathrm{~J}=1.0 \mathrm{~J} $$ Can you show that the order in which the charges are brought in from infinity does not affect this result?

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