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zysi [14]
3 years ago
7

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in

side and outside the ball is the magnitude of the ball's electric field equal to 1 3 of the maximum magnitude of that field? (a) inside the ball R (b) outside the ball R

Physics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where  \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

Read more on Brainly.com - brainly.com/question/13242041#readmore

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A 60 [Hz] high-voltagepower line carries a current of 1000 [A]. The power line is at a height of 50 [m] above the earth. What is
omeli [17]

Answer:

The magnitude of the magneticfield B[T] at a point on the surface of the earth directly below the power line is B=1.496x10^(-6) T.

Explanation:

The distance from the wire to a point in the surface is the heigth of the wire.

The formula for the magnetic field on any point at distance R from a wire conducting alternating current is:

B=\frac{\mu_0I}{2\pi R} =\frac{(4\pi\cdot 10^{-7}T\cdot m/A)(1000\,A)}{2\pi \cdot 50\,m} =1.496\cdot10^{-6}\,T

4 0
3 years ago
How density of substance change with change in temperature?
Anna71 [15]

The answer is above but I don't know if it's correct.

7 0
2 years ago
Why isn’t our constitution ( and our country) in a constant state of adjustment and readjustment
madam [21]

The constitution and the country aren't in a constant state of adjustment and readjustment because it has strong and logical ideological bases that encompass most of society's issues over time.

The constitution was written on a democratic ideological basis inspired by the European Enlightenment revolution of thought. These principles have contributed to the construction of American social logic.

Therefore, the constitution and the country have not needed major readjustments or adjustments since their creation because the country has been built on solid and well-defined ideological bases that are projected into the future with a social logic.

Learn more in: brainly.com/question/14453917

4 0
3 years ago
3. If atoms are mostly empty space, why can't<br> we walk through walls?
Serhud [2]

Answer:

Good question

Explanation:

The reason is that walls are made up of more than just atoms. Walls are built up from many things, so with this being said we cannot walk through the walls

PPs why do or should anyone want to walk through walls?

7 0
3 years ago
A garden hose with a small puncture is stretched horizontally along the ground. The hose is attached to an open water faucet at
tresset_1 [31]

Answer:

The pressure inside the hose 7000 Pa to the nearest 1000 Pa.

<em>h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}</em>

<em>V_{1} =V_{2}</em>

The pressure at the site of the puncture is <em>P₂ =7161.3 Pa</em>

Explanation:

<em>According to Poiseuille's law, P_{1} - P_{2}  = \frac{128μLQ}{πD^{4} } </em>

<em>Where P_{1} is the pressure at a point 1 before the leak, P_{2} is the pressure at the point of  the leak 2, μ = dynamic viscosity, L = the distance between points 1 and 2, Q = flow rate, D = the diameter of the garden hose. </em>

<em>Also,  from the equation P =ρgh, the equations h₁ = \frac{P₁} {ρg} and h₂ = \frac{P₂} {ρg} can be derived.</em>

Combining Poseuille's law with the above, we get h₁ - ρgh₂ =  \frac{128μLQ}{πD^{4} }

<em>h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}</em>

<em>V =\frac{Q}{A}</em>

Since the hose has a uniform diameter, the nozzle at the end is closed and neither point <em>1 nor 2 lie after the puncture,</em>

<em>V_{1} =V_{2}</em>

The pressure at the site of the puncture <em>P₂ =ρgh₂</em>

<em>P₂ =1kgm⁻³ ×9.81ms⁻¹×0.73m</em>

<em>P₂ =7161.3 Pa</em>

<em />

7 0
3 years ago
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