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Mumz [18]
3 years ago
14

Consider two thin disks, of negligible thickness, of radiusR oriented perpendicular to thex axis such that the x axis runs throu

gh thecenter of each disk. The disk centered at x=0 has positive charge density eta, and the disk centered at x=a has negative charge density -\eta, where the charge density is charge perunit area. What is the magnitude E of the electric field at the point on thex axis with x coordinate a/2? For what value of the ratio R/a of plate radius to separation between the plates does the electric field at the point x=a/2 on the x axis differ by 1 percent from the result η/ϵ for infinite sheets?

Physics
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Complete question

The complete question is shown on the first and second uploaded image

Answer:∈

Answer to first question is shown on the second uploaded image.

Part B the Answer is:

The ratio  \frac{R}{a} is evaluated to be 49.99

Explanation:

The explanation is shown on the third ,fourth and fifth image.

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Which kinds of objects emit visible light in the electromagnetic spectrum?
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Answer : Relatively hot objects

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We know the range of wavelength of the visible spectrum is from 400 nm to 780 nm.





3 0
3 years ago
Read 2 more answers
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va
WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V =  3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0
3 years ago
How does gas exert pressure? Explain using the change in momentum.
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Answer:

The molecules are continually colliding with each other and with the walls of the container. When a molecule collides with the wall, they exert small force on the wall The pressure exerted by the gas is due to the sum of all these collision forces. The more particles that hit the walls, the higher the pressure.

4 0
3 years ago
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A box with a mass of 100.0 kg slides down a ramp with
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Answer:

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I got the answer right

3 0
3 years ago
Part a of the drawing shows a bucket of water suspended from the pulley of a well; the tension in the rope is 90.5 n. part b sho
Butoxors [25]
Missing figure in the problem: https://www.physicsforums.com/attachments/p4-46-gif.117834/

Solution:

part a) The bucket is not moving so the resultant of the forces acting on it is zero. Let's apply the equation of equilibrium. We have:
mg-2T=0
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mg=2\cdot 90.5 N= 181 N

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mg-T=0
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T=mg=181 N


3 0
3 years ago
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