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Nesterboy [21]
2 years ago
12

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

e xy plane.
What is the electric potential V(z) on the z axis as a function of z , for z>0 ?
What is the magnitude E of the electric field on the z axis, as a function of z , for z>0 ?
Physics
1 answer:
sleet_krkn [62]2 years ago
7 0

The electric potential V(z) on the z-axis is :  V = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z  

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  \frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = \int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,

 V = \pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]

     = \pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} }  -z ]

Therefore the electric potential V(z) = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

Also

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

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Express force in terms of base units​
MariettaO [177]
<h2><u>Answer:</u></h2>

<u></u>

<u>Force = kgm/s² ⟶ Newton</u>

<h2><u>Explanation:</u></h2>

<u></u>

According to the formula we've learnt,

  • Force = mass × acceleration.

To find force in terms of base units, we must first identify the base SI units of mass & acceleration.

  • Base SI unit of mass = kg (kilogram)
  • Base SI unit of acceleration = m/s² (meter per second squared)

So, the base unit of force is,

  • Force = mass × acceleration.
  • Force = kg × m/s²
  • <u>Force = kgm/s² ⟶ Newton</u>

________________

Hope it helps!

\mathfrak{Lucazz}

4 0
2 years ago
Find the first three harmonics of a string of linear mass density 2. 00 g/m and length 0. 600 m when the tension in it is 50. 0
gavmur [86]

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

<h3>Velocity of the wave</h3>

The velocity of the wave is calculated as follows;

v = √T/μ

where;

  • T is tension
  • μ is mass per unit length = 2 g/m = 0.002 kg/m

v = √(50/0.002)

v = 158.1 m/s

<h3>First harmonic or fundamental frequency of the wave</h3>

f₀ = v/λ

where;

  • λ is the wavelength = 2L

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

<h3>Second harmonic of the wave</h3>

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

<h3>Third harmonic of the wave</h3>

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

Learn more about harmonics here: brainly.com/question/4290297

#SPJ1

6 0
1 year ago
Please help me with the question below the best answer with an explanation will get brainliest
lesya [120]

Answer:

B

Explanation:

i hope this is correct.

6 0
2 years ago
One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms. Assume that
Rufina [12.5K]

Answer:

45930.52N

Explanation:

Net force = (internal pressure - external pressure)× area of window

Net force = (1.02 - 0.910)atm × 2.03m × 2.03m = 0.11atm × 4.1209m^2 = 0.11 × 101325N/m^2 × 4.1209m^2 = 45930.52N

5 0
3 years ago
Which is the result of using a machine?
Eddi Din [679]
Correct Answers is A.

The machines gives us some mechanical advantage. This means the mechanical average makes the work output greater than the work input
Simple most example is a lever. The force applied is smaller and the output work is larger as compared to input.

Option B cannot be true, as there must be a force to get some work done.
Option C and D are inverse of what a machine is designed for. A small force can be exerted through a large distance to have a large force exerted through a small distance. Common Example of this principle is a screw opener. 
3 0
3 years ago
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