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1 year ago

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A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

2R that also carries charge Q. The charge Q is distributed uniformly over the insulating shell.Find the magnitude of the electric field in the region 02R. Express your answer in terms of the variables R, r, Q, and constants π and ε0.

1 answer:

Nostrana [21]1 year ago

4 0

Explanation:

Gauss Law relates the distribution of electric charge to the resulting electric field.

Applying Gauss's Law,

EA = Q / ε₀

Where:

E is the magnitude of the electric field,

A is the cross-sectional area of the conducting sphere,

Q is the positive charge

ε₀ is the permittivity

We be considering cases for the specified regions.

<u>Case 1</u>: When r < R

The electric field is zero, since the enclosed charge is equal to zero

E(r) = 0

<u>Case 2</u>: When R < r < 2R

The enclosed charge equals to Q, then the electric field equals;

E(4πr²) = Q / ε₀

E = Q / 4πε₀r²

E = KQ /r²

Constant K = 1 / 4πε₀ = 9.0 × 10⁹ Nm²/C²

<u>Case 3</u>: When r > 2R

The enclosed charge equals to Q, then the electric field equals;

E(4πr²) = 2Q / ε₀

E = 2Q / 4πε₀r²

E = 2KQ /r²

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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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You need to determine the density of an unknown liquid and decide to perform an experiment. You notice that a wooden block float

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Answer:

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

Explanation:

The density of a liquid is inversely proportional to the volume (height) of object submerged in it.

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p ∝ 1/V ∝ 1/h

since V = Ah

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p = density

h = height submerged

pu and pw is the density of unknown liquid and water respectively

hu and hw is the height of object submerged in unknown liquid and water respectively

pw = 1000kg/m^3

hu = 4.6cm = 0.046m

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Substituting the given values;

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Answer:

Explanation:

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mv - mu , v and u are final and initial velocity during impact at surface

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v² = u² + 2gh₁

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v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

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Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

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We need to calculate the spring constant

Using formula of mechanical energy of the system

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2$

$k=\dfrac{2.5\times2}{(10\times10^{-2})^2}$

$k=500\ N/m$

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

$E=15\ J$

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Now, we need to find the direction the bee should fly directly to the flower (due North):

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Clearing $\theta$ :

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