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3 years ago

6

How does the configuration of the electric field occur between a "parallel plate" setup in a lab?What is the effect of conductor

s (rings and the triangle plates) on the electric field and equipotential lines?

1 answer:

sleet_krkn [62]3 years ago

8 0

Explanation:

The magnitude of the electric field between the plates is given by

E = -ΔV/d

minus sign indicates Potential decreases in the direction of electric field

where

ΔV is the potential difference between the plates

D is the distance between the plates.

The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.

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igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

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Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

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Answer:

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Explanation:

We are given that

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Initial speed, u=1.4 m/s

Acceleration, $a=0.20 m/s^2$

We have to find the final velocity at the end of the 100.0 m.

We know that

$v^2-u^2=2as$

Using the formula

$v^2-(1.4)^2=2\times 0.20\times 100$

$v^2-1.96=40$

$v^2=40+1.96$

$v^2=41.96$

$v=\sqrt{41.96}$

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A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

Paha777 [63]

Answer:

c. V = 2 m/s

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so:

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Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = $\frac{1}{2}MR^2$

I = $\frac{1}{2}(5kg)(2m)^2$

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so, we can write the initial equation as:

Mgh = $\frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2$

Replacing the data:

(5kg)(9.8)(0.3m) = $\frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2$

solving for V:

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