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ra1l [238]
3 years ago
8

the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located

2.00 mm above the dipole's midpoint

Physics
1 answer:
vichka [17]3 years ago
3 0

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

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The work output is 3.02\cdot 10^5 J

Explanation:

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Keeping in mind that

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3 years ago
A particular car has a weight of 9500 N. What is the mass of the car? 9500 kg 95000 kg 950 kg 2160 kg
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Pendulum mass is 4 kg. Use your equations for gravitational potential energy and kinetic energy to determine these values based
mrs_skeptik [129]

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How much heat in kilojoules is required to melt an ice cube with a mass of 18.6 g at 0 °C? The Lf for water is 333 J/g.
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Hope this helps!



3 0
3 years ago
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