1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mylen [45]
3 years ago
5

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

ductor (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.
Physics
1 answer:
Tcecarenko [31]3 years ago
5 0

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

You might be interested in
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th
Igoryamba

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

5 0
3 years ago
The pressure inside a sealed container of methane gas (CH4) is 35.0 kPa. If this 80.0 L sample
STatiana [176]

Answer:

<h3>The answer is 30.43 L</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{35000 \times 80}{92000}  =  \frac{2800000}{92000}  =   \frac{700}{23}  \\  = 30.434782...

We have the final answer as

<h3>30.43 L</h3>

Hope this helps you

4 0
3 years ago
A 0.10 kg ball of dough is thrown straight up into the air with an initial speed of 15 m/s.
Reptile [31]
<span>Mass of the ball is m = 0.10kg Initial speed of the Ball v = 15m/s a. When the ball is at maximum height the velocity is 0 Momentum of ball = mass x velocity Momentum = 0.10kg x 0 = 0 b. Getting the maximum height, Using the conservation of energy equation KEinitial = mgh 1/2mVin^2 = mgh => h = v^2/2g h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m Applying the conservation of energy equation at halfway V^2 = 2gh V = square root of (2x9.8x5.96) => V = square root of (116.816) So the velocity at the half way V = 10.81 m/s Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>
6 0
3 years ago
There. That is better.
mihalych1998 [28]

a
a
b
b
a
b
a
This will really help you learn a lot.

6 0
3 years ago
Where can radiation be found in nature and how is it affected
Murljashka [212]
The sun is a clear example of objects releasing radiation in nature
8 0
2 years ago
Other questions:
  • What is the surface area to volume ratio of this cube
    6·1 answer
  • “doing the wave” is a common activity in large football stadiums. what type of wave is this?
    5·1 answer
  • How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 j/g °c to increas
    6·1 answer
  • I need to find 1).a,b,c
    15·1 answer
  • Wonder Woman and Superman fly to an altitude of 1690 km , carrying between them a chest full of jewels that they intend to put i
    10·1 answer
  • Which pair of elements is most similar?<br> Na and CI<br> Li and Ne<br> Co and<br> Ne and Ar
    5·1 answer
  • A small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. Which set of vectors gives the di
    9·1 answer
  • Two car horns are sounded creating two sound waves with frequencies that differ by a factor of three. How does the speed of the
    10·1 answer
  • A 195 g glider is moving at 5.3 m/s on a frictionless air track. It then collides with a stationary 295 g glider.
    5·1 answer
  • A single-phase 60-Hz overhead power line is symmetrically supported on a horizontal cross arm. Spacing between the centers of th
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!