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grin007 [14]
3 years ago
6

An infinite line charge of linear density λ = 0.30 μC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis a

t y = 2.0 m. The x component of the electric field at the point P on the x axis at x = 3.0 m is approximately . A) 1.8 kN/C B) 4.2 kN/C C) 0.96 kN/C D) 5.2 kN/c E) 0.64 mN/C
Physics
2 answers:
ira [324]3 years ago
6 0
The line charge E-field Ec = λ/(2πr*e0),
where λ = charge/length and e0 is the permittivity constant = 8.8542E-12 F/m. 
<span>The point charge E-field Ep = kq/r^2 where electrostatic constant k = 1/(4π*e0) = 8.99E9 N-m^2/C^2.</span>
Ivahew [28]3 years ago
6 0

Answer:

D) 5.2 kN/c

Explanation:

Electric field due to infinity line charge

λ /2πε₀r = .3 x 10⁻⁶ x 2 x 9 x 10⁹ / 3 = 1.8 x 10³. It will be towards x axis.

Electric field due to point charge = k.q / r² = 9 x 10⁹ x 6 x 10⁻⁶ / 13 = 4.15 x 10³

x component of this field = 4.15 x 10³ x cos Ф = 4.15 x 10³ x 3 / √13

3.458 x 10³

Total electric field = 5.2 kN / C.

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You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do
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Answer:

Work done in both the cases will be same

Explanation:

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here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

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Nadusha1986 [10]
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8 0
3 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

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Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

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6 0
4 years ago
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