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scoundrel [369]
4 years ago
5

A circular loop of radius R = 3 cm is centered at the origin in the region of a uniform, constant electric field. When the norma

l to the loop points in the positive x-direction the flux through the loop is Φ1 = 85 N•m2/C. When the normal to the loop points in the positive y-direction the flux through the loop is Φ2 = -85 N•m2/C. When the normal to the loop points in the positive z-direction the flux through the loop is zero.
Physics
1 answer:
hram777 [196]4 years ago
6 0

Answer:

The Questions are

a. Calculate the x-component of the electric field, in newtons per coulomb

b. Calculate the y-component of the electric field, in newtons per coulomb

c. Calculate the z-component of the electric field, in newtons per coulomb

d. Calculate the magnitude of the electric field, in newtons per coulomb.

Explanation:

Given that,

Φx= 85 N•m2/C. X direction

Φy= -85 N•m2/C. Y direction

Φz = 0. Z direction

Radius of loop =3cm=0.03m

Surface area of the circle is πr²

A=22/7×0.03²

A=0.00283m²

Flux is given as Φ=EA

a. Φx=ExA

Ex=Φx/A

Ex=85/0.00283

Ex=30035.33N/C

b. Ey=Φy/A

Ey=-85/0.00283

Ey=-30035.33N/C.

c. Ez=Φz/A

Ez, =0/A

Ez=0N/C

d. Magnitude of E.

E=√Ex²+Ey²+Ez²

E=√(30035.33)²+(-30035.33)²

E=42476.38N/C.

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Can someone solve this problem and explain to me how you got it​
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1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

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r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

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Therefore, the new force between the charges is:

F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

F=k\frac{q_1 q_2}{r^2}

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

q_1' = 8q_1

q_2' = 8q_2

Substituting into the equation, we find the new force:

F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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