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lianna [129]
3 years ago
6

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

ion for E=kQ/r2 for r>R and E=kQr/R3 for r
Only need C & D. It asks for specfic points that I don't know how to find. I just know the general shape of the graph.
Physics
1 answer:
AlladinOne [14]3 years ago
8 0

Answer:

Vb = k Q / r        r <R

Vb = k q / R³ (R² - r²)    r >R

Explanation:

The electic potential is defined by

             ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

             VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

           Va = - ∫ (k Q / r²) dr

           -Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

          Vb = k Q / r        r <R

         

We perform the calculation of the power with the expression of the electric field that they give us

           Vb = - int (kQ / R3 r) dr

  We integrate and evaluate from the starting point r = R to the final point r <R

         Vb = ∫kq / R³ r dr

         Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

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A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th
sleet_krkn [62]

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

7 0
2 years ago
Find the equation of the line below.
Musya8 [376]

Answers all in picture below

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7 0
3 years ago
Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec
yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

mg=kx

k=\dfrac{mg}{x}

k=\dfrac{3.74\times 9.8}{0.0161}

k = 2276.52 N/m

The frequency of vibration is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

8 0
3 years ago
Gggcdfubrdegubtcwftvf y day rx u e HHS’s forget h
aev [14]

Answer:

um . . . yes ?

8 0
2 years ago
Read 2 more answers
a uniform rule balance on a knife edge at the 55cm mark when a mass of 40g is hung from the 95cm mark. find the weight of the ru
BigorU [14]

Answer:

W = 3.1 N

Explanation:

moments about any convenient point will sum to zero.

I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.

I will assume the ruler weight makes a positive moment

W[55 - 50) - 0.040(9.8)[ 95 - 55] = 0

                                                5W = 15.68

                                                   W = 3.136

7 0
2 years ago
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