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AP Physics 2 : Capacitors and Electric Fields

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #301 : Electricity And Magnetism

A capacitor is placed in series with a $\displaystyle 6V$battery. The plates are 1mm $\displaystyle 1mm$ apart and each have a 4cm^2 $\displaystyle 4cm^2$ surface area. Determine the charge on the positive plate.

Possible Answers:

$4.54*10^{-11}C$ $\displaystyle 4.54*10^{-11}C$

$3.95*10^{-11}C$ $\displaystyle 3.95*10^{-11}C$

$2.13*10^{-11}C$ $\displaystyle 2.13*10^{-11}C$

$8.44*10^{-11}C$ $\displaystyle 8.44*10^{-11}C$

None of these

Correct answer:

$2.13*10^{-11}C$ $\displaystyle 2.13*10^{-11}C$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combining equations:

V_1=E*d $\displaystyle V_1=E*d$

Solving for $\displaystyle E$:

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

Converting $\displaystyle mm$ to $\displaystyle m$ and plugging in values:

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

$6000\frac{N}{C}=E$ $\displaystyle 6000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$ $\displaystyle E=\frac{q}{A*\epsilon_0}$

Combining equations:

$E*\epsilon_0*A=q$ $\displaystyle E*\epsilon_0*A=q$

Converting cm^2 $\displaystyle cm^2$ to m^2 $\displaystyle m^2$ and plugging in values:

$(6000)*(8.86*10^{-12})*(4*10^{-4})=q$ $\displaystyle (6000)*(8.86*10^{-12})*(4*10^{-4})=q$

$q=2.13*10^{-11}C$ $\displaystyle q=2.13*10^{-11}C$

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Example Question #311 : Electricity And Magnetism

A capacitor is placed in series with three parallel $\displaystyle 6V$batteries. The plates are 1.5mm $\displaystyle 1.5mm$ apart and each have a 10cm^2 $\displaystyle 10cm^2$ surface area. Determine the charge on the positive plate.

Possible Answers:

$8.44*10^{-11}C$ $\displaystyle 8.44*10^{-11}C$

None of these

$3.08*10^{-11}C$ $\displaystyle 3.08*10^{-11}C$

$5.55*10^{-11}C$ $\displaystyle 5.55*10^{-11}C$

$1.42*10^{-11}C$ $\displaystyle 1.42*10^{-11}C$

Correct answer:

$1.42*10^{-11}C$ $\displaystyle 1.42*10^{-11}C$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor. Since the batteries all have the same potential and are in parallel, the potential will be the same as if there was just one.

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combining equations:

V_1=E*d $\displaystyle V_1=E*d$

Solving for E:

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

Converting $\displaystyle mm$ to $\displaystyle m$ and plugging in values:

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

$4000\frac{N}{C}=E$ $\displaystyle 4000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$ $\displaystyle E=\frac{q}{A*\epsilon_0}$

Combining equations:

$E*\epsilon_0*A=q$ $\displaystyle E*\epsilon_0*A=q$

Converting cm^2 $\displaystyle cm^2$ to m^2 $\displaystyle m^2$ and plugging in values:

$(4000)*(8.86*10^{-12})*(4*10^{-4})=q$ $\displaystyle (4000)*(8.86*10^{-12})*(4*10^{-4})=q$

$q=1.42*10^{-11}C$ $\displaystyle q=1.42*10^{-11}C$

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All AP Physics 2 Resources

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AP Physics 2 : Capacitors and Electric Fields

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #301 : Electricity And Magnetism

Example Question #311 : Electricity And Magnetism

All AP Physics 2 Resources

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