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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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A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to
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Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV  

V = W/q    ....(1)

Using equation (1), the potential difference between points A and B is:

V_{1}=\frac{W_{1} }{q}

Substitute the suitable values in the above equation.

V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

V_{2}=\frac{W_{2} }{q}

Substitute the suitable values in the above equation.

V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0
3 years ago
Why is the combination of two protons and two neutrons stable, but two protons and one neutron is not?
RideAnS [48]
Because it's unbalanced.
4 0
2 years ago
Describe what happens to the particles of water when it boils.
Nezavi [6.7K]

Answer:

The liquid turns to a gas.

Explanation:

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car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a
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The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

  • <em>Length of the road, L = 320 km</em>
  • <em>Distance covered = 240 km at 75 km/h</em>
  • <em>time spent refueling, t₂ = 0.6 hr</em>
  • <em>Final velocity, = 100 km/hr</em>

The time spent by the before refueling is calculated as follows;

t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours

The time spent by the car for the remaining journey;

t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr

The total time of the journey is calculated as follows;

t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours

The average velocity of the car for the whole journey is calculated as follows;

v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h

Learn more about average velocity here: brainly.com/question/6504879

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Answer:

I don't know the answer I need points before I fail my math test

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2 years ago
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