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Lisa [10]
1 year ago
10

A straight, cylindrical wire lying along the x axis has a length L and a diameter d . It is made of a material described by Ohm'

s law with a resistivity rho . Assume potential V is maintained at the left end of the wire at x = 0. Also assume the potential is zero at x=L . In terms of L, d , V, rho, and physical constants, derive expressions for (a) the magnitude and direction of the electric field in the wire,
Physics
1 answer:
Zepler [3.9K]1 year ago
7 0

The magnitude and direction of the electric field in the wire are mathematically given as

L &=[(v / L) v / m] \hat{i}

<h3>What is the magnitude and direction of the electric field in the wire?</h3>

Generally, the equation for is  mathematically given as

A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides  

E d x=\int d v.

\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}

In conclusion, the magnitude and direction of the electric field in the wire are given as

L &=[(v / L) v / m]

Read more about electric fields

brainly.com/question/15800304

#SPJ4

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A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

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Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

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8 0
3 years ago
Use Hooke's Law, which states that the distance a spring stretches (or compresses) from its natural, or equilibrium, length vari
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Answer:

706.68 N

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F = ke

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Using the values in the question,

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When e = 0.4 m,

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Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be  

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