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Law Incorporation [45]
3 years ago
15

You have now seen examples during lecture on how to calculate the electric field for a line of charge and a ring of charge - bot

h uniformly distributed. This activity will ask you to solve for the electric field, on-axis, of a uniformly-charged disk sitting in the yz plane. Below is a picture of the situation of interest. Note: Treat it as a totally flat disk and ignore its thickness in the x direction. Also, let x be the distance between the center of the disc and point P.
Physics
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

1/4πε₀[Hx/ (√x² + b²)^3/2]i.

Explanation:

So, without mincing words let's dive straight into the solution to the question above. There is need to determine the electric field on-axis of a uniformly-charged disk sitting.

The electric field in the x-component, dεₓ = 1/4πε₀[H/ x² + b²] cos .

Thus, the total electric field in the x-component, εₓ = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2.

Therefore, the electric field = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2i.

Where x=0

You might be interested in
A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?
Ilya [14]

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

                  = 64°C

<u>And now we have to write the temperature in kelvins.</u>

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

<u>Let us find it now.</u>

64°C + 273 = 337k

Therefore,

64°C ⇒ <u>337k</u>

8 0
1 year ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
2 years ago
11.
jolli1 [7]
<h3>Reducing Surface Area.</h3>

If the surface area becomes smaller, the pressure becomes larger.

8 0
3 years ago
Please help on this one?
kupik [55]

Answer: A)30V. First find the current of the circuit. I=V/R(total resistance). So I=60/120=0.5. Now to find voltage drop in R3 use ohms law as given. V(of 3)=(0.5)(60)=30V

7 0
2 years ago
An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0
2 years ago
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