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3 years ago

Please help me TT. I need this to be submitted soon TT. Thank you

Physics

1 answer:

Verizon [17]3 years ago

3 0

Answer:

a) f = 1m × 2 × (5A / √2) × (5A / √2) / 0.003m = 0.00166... (66 is repeating)

b) The currents on two wires on a AC chord are always moving in opposite direction and so they are always replusing.

c) There needs to be a sheath to dampen the replusing, fluctuating force of the wires.

a) v = √( ( (-2)(-1.6 × 10^(-16))(3000V) ) / (2.84 × 10^(-20)kg) ) = 5.81227 × 10^3

b) Any ion transversing a chamber having a magnetic field will deflect.

c) The direction of the electric field is vertical because it's perpendicular to the plates. The electric field magnitude is independent from the magnitude of the magnetic field and charge. So it's not possible to find the magnitude of the electric field, without knowing the voltage on the plates, the distance between the plates, and the dielectric constant.

d) Assuming the mangetic field remained, the path of the negative ions will be deflected vertically given that the magnetic field is horizontally perpendicular to the negative charged ions movement.

Sorry it took so long :) If anything is incorrect please let me know.

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3 years ago

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Answer:

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2 years ago

What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?

IRISSAK [1]

Answer:

The magnitude of electric field is 22.58 N/C

Solution:

Given:

Force exerted in upward direction, $\vec{F_{up}} = 3.50\times 10^{- 5} N$

Charge, Q = $- 1.55\micro C = - 1.55\times 10^{- 6} C$

Now, we know by Coulomb's law,

$F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}$

Also,

Electric field, $E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}$

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

$\vec{E} = \frac{\vec F_{up}}{Q}$

$\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}$

$\vec{E} = - 22.58 N/C$

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

$|\vec{E}| = 22.58\ N/C$

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3 years ago

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so, 1.02 × 10^(-13)

so the most appropriate answer is 2 ...that is
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4 years ago

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3 years ago

Please help me TT. I need this to be submitted soon TT. Thank you​

Please help me TT. I need this to be submitted soon TT. Thank you