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3 months ago

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An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the

electric field outside the cylinder is Er=λ2πϵ0r. Part A Find the potential relative to the surface at a point that is distance r from the axis, assuming r>R. Express your answer in terms of the variables λ, r, R, V0, and appropriate constants. Vr = nothing

1 answer:

kicyunya [3.2K]3 months ago

5 0

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

The complete solution is provided in the attachment.

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The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

kicyunya [3294]

To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

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3 months ago

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If 2.0 mol of gas a is mixed with 1.0 mol of gas b to give a total pressure of 1.6 atm, what is the partial pressure of gas a an

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<span>The partial pressure of A = 1.06 atm and the partial pressure of B = 0.53 atm</span>

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2 months ago

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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

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Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Distance = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

The potential difference is calculated as

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference across the plates amounts to 10000 V

Area is determined by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$

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2 months ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

serg [3582]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: The speed of a wave on a string under tension can be determined using the following:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

Distance a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5× $10^{-6}$ m.

When tension is doubled:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Distance in the same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With the increased tension, it moves $\Delta x =$ 15.4× $10^{-5}$ m

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