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polet [3.4K]
3 years ago
7

Need help on question b

Physics
1 answer:
ZanzabumX [31]3 years ago
8 0

The answer is A.

A positive charge’s electric field pushes out.

Hope this helps! -Avenging

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
Elaborate on the classification of aluminum. A) Aluminum is a noble gas and has a full valence electron shell, is chemically non
scZoUnD [109]
The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
</span><span>
</span>
4 0
3 years ago
Read 2 more answers
cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?
serg [7]

The heat moves from the hot chocolate to the handle of the spoon by a process called thermal conduction.

It is the transfer of heat energy from one object to another when they are in contact with eachother.

Hope this answers your question.

5 0
3 years ago
While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far
viktelen [127]

Answer:

1.7 m

Explanation:

v_x = Velocity of ball in x direction = 4.47 m/s

u_y = Velocity of ball in y direction = 0

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

t = Time taken

s_y = Vertical displacement = 0.7 m

s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}

Horizontal displacement is given by

s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}

The passenger should throw the ball 1.7 m in front of the bucket.

5 0
3 years ago
When frequency is decreased what happens to the amplitude?
guajiro [1.7K]

Answer:

it increases the amplitude of the wave as it propagates.

Explanation:

7 0
2 years ago
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