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mihalych1998 [28]
4 years ago
11

An electric dipole is placed in a uniform electric field. Which options below are true? [Enter your answer as a string of the le

tter options you believe are correct. For instance, if you think options A, D, and E are correct, then enter "ADE"] A) The dipole is never in equilibrium since it will always twist around. B) A dipole whose moment vector is not aligned with a uniform electric field will twist around toward being aligned. C) No net force acts on the electric dipole. D) There are no electric forces exerted on each pole of the dipole. E) None of the above
Physics
1 answer:
slega [8]4 years ago
6 0

Answer:

C and D are correct

Explanation:

An electric dipole moment for pair of equal and opposite charges is defines the magnitude of charge times the distance between them and the direction is always towards the positive charge.  According to the definition only points C and D are correct.

C) No net force acts on the electric dipole.

D) There are no electric forces exerted on each pole of the dipole.

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Answer:

a) Speed of the particle at this instant

v = 8.43 m/s

b) Speed of the particle at  (1/8) revolution later

v = 14.83 m/s

Explanation:

We apply the equations of circular motion uniformly accelerated :

(a_{T}) ^{2} = (a_{n} )^{2} +(a_{t} )^{2} Formula (1)

a_{n} = \frac{v^{2} }{r} Formula (2)

a_{t} = \alpha *r Formula (3)

v= ω*r Formula (4)

ω² = ω₀² + 2*α*θ  Formula (5)

Where:

a_{T} :  total acceleration, (m/s²)

a_{n} : normal acceleration, (m/s²)

a_{t} :  tangential acceleration, (m/s²)

\alpha : angular acceleration (rad/s²)

r : radius of the circular path (m)

v : tangential velocity (m/s)

ω : angular speed ( rad/s)

ω₀: initial angular speed  ( rad/s)

θ : angle that the particle travels (rad)

Data:

a_{T} = 15 m/s²

a_{t} =  12 m/s²

r=7.90 m  :radius of the circular path

Problem development

In the formula (1) :

a_{n} = \sqrt{(a_{T})^{2} -(a_{t})^{2} }

We replace the data

a_{n} = \sqrt{(15)^{2} -(12)^{2}}

a_{n} = 9 \frac{m}{s^{2} }

We use formula (2)  to calculate v:

9 = \frac{v^{2} }{7.9}  Equation (1)

a)Speed of the particle at this instant

in the equation (1):

v=\sqrt{9*7.9} = 8.43 \frac{m}{s}

b)Speed of the particle at  (1/8) revolution later

We know the following data:

θ =(1/8) revolution=( 1/8) *2π= π/4

a_{t} =  12 m/s²

v₀= 8.43 m/s

r=7.9 m

We use formula (3) to calculate α

12 = \alpha *7.90

\alpha =\frac{12}{7.9} = 1.52  \frac{rad}{s^{2} }

We use formula (4) to calculate ω₀

v₀= ω₀ *r

8.43 =  ω₀*7.9

ω₀ = 8.43/7.9 = 1.067 rad/s

We use formula (5) to calculate ω

ω² = ω₀² +  2*α*θ  

ω²=  (1.067)² + 2*1.52*π/4

ω² =3.526

ω = 1.87 rad/s

We use formula (4) to calculate v

v= 1.87 rad/s * 7.9m

v = 14.83 m/s : speed of the particle at  (1/8) revolution later

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