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1 year ago

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The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)

1.69y2 +15.6y 52.5) Volts. Calculate the magnitude of the electric field at the position y - 22.5 meters

1 answer:

nikdorinn [45]1 year ago

3 0

Answer:

$E = 55.9583\ Volts/meter$

Explanation:

First let's find the electric potential using y = 22.5:

$V(y) = 1.69y^2 +15.6y+52.5$

$V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5$

$V(22.5) = 1259.0625\ Volts$

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

$E = V/d$

$E = 1259.0625/22.5$

$E = 55.9583\ Volts/meter$

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Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12

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$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

$-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-$

In case all three capacitors are connected in parallel, the $2.0\;\mu\text{F}$ capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

<h3>Case B: 5.54 V.</h3>

$-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-$

In case the $2.0\;\mu\text{F}$ capacitor is connected in parallel with the $1.5\;\mu\text{F}$ capacitor, and the two capacitors in parallel is connected to the $3.0\;\mu\text{F}$ capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}$ .

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

$Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}$ .

Voltage is the same across two capacitors in parallel.As a result,

$\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}$ .

<h3>Case C: 2.76 V.</h3>

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

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the effective capacitance of the two capacitors in parallel is 5.0 μF;
the effective capacitance of the three capacitors, combined: $\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}$ .

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<h3 /><h3>Case D: 4.00 V</h3>

$-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-$ .

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Read 2 more answers

A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency

allochka39001 [22]

Explanation:

It is given that,

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(1) The frequency of the wave is given by :

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$\nu=\dfrac{251\ m/s}{0.051\ m}$

$\nu=4921.56\ Hz$

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$\omega=2\pi\nu$

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(3) The period of oscillation is given by T as :

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$T=\dfrac{1}{4921.56}$

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

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