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Dahasolnce [82]

2 years ago

14

We find that the electric field of a charged disk approaches that of a charged particle for distances y that are large compared

to R, the radius of the disk. To see a numerical instance of this, calculate the magnitude of the electric field a distance y = 2.9 m from a disk of radius R = 2.9 cm that has a total charge of 5.0 µC using the exact formula as follows. (Enter your answer to at least one decimal place.)
E = 2πK σ(1- [y/(R² + y²)]) j^

1 answer:

Alex2 years ago

8 0

Answer:

7. 01 * 10^7 N/C

Explanation:

Parameters given:

Distance, y = 2.9 m

Radius, R = 2.9 cm = 0.029m

Charge, Q = 5.0 µC = 5 * 10^(-6) C

Given that:

E = 2πKσ(1- [y/(R² + y²)]) j^

Charge density, σ, is given as:

σ = Q/A = Q/πR²

=> E = 2πKQ/πR² (1- [y/(R² + y²)]) j^

E = 2KQ/R² (1 - [y/(R² + y²)]) j^

E = (2 * 9 * 10^9 * 5 * 10^(-6) / 0.029²) (1 - [2.9/(0.029² + 2.9²)]) j^

E = 107015.46 * 10^3 *(1 - 2.9/8.41) j^

E = 107015.46 * 10^3 * 0.655 j^

E = 7.01 * 10^7 j^ N/C

The magnitude of the electric field is 7.01 * 10^7 N/C. The j^ shows the direction.

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A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

kari74 [83]

Answer:

$a. 4.2057\times 10^-^1^2 F \ or 4.2057\ pF\\b. 2.7344V$

Explanation:

a.

Given the permittivity constant to be $8.854\times 10^-^1^2 F/m$ ,The capacitance of a $cylindrical \ capacitor$ of length, $L$ is given by the equation:

$C=\frac{2\pi \epsilon _o L}{ln(b/a)}$ where $b$ is the radius of the outer cylinder and $a$ the radius of the inner cylinder.

The values are given as: $a=0.550mm(5.5\times 10^-^4m), \ b=4.00mm(4.0\times10^-^3m), \ L=15.0cm(0.150m)$

Substitute in our capacitance equation:

$C=\frac{2\times\pi \times 8.854\times 10^-^1^2 \times 0.15}{In(4.00/0.550)}\\=4.2057\times 10^-^1^2 F$

Hence the capacitance is $4.2057\times 10^-^1^2 F$

b. The charge on the capacitance is related to the potential difference across it. The potential difference is expressed using the equation:

$Q=CV$ , $Q=11.5pC$

From a above, we already have our capacitance value, $C=4.2057\times 10^-^1^2 F$

We substitute $C$ in the pd equation:

$v=>(11.5)/(4.2057)\\=2.7344V$

Hence, the applied potential difference is 2.7344V

4 0

3 years ago

What is pulling force expansion

Aloiza [94]

Answer:

physical cosmology, the Big Rip is a hypothetical cosmological model concerning the ultimate fate of the universe, in which the matter of the universe, from stars and galaxies to atoms and subatomic particles, and even spacetime itself, is progressively torn apart by the expansion of the universe at a certain time

3 0

2 years ago

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HELP!!! What role might electrostatic force play in spider dispersal, according to a recent study?

Marat540 [252]

The positively charged atmosphere attracts negatively charged spider silk, might electrostatic force play in spider dispersal, according to a recent study.

Answer: Option C

<u>Explanation:</u>

The positive charge present in upper of the atmosphere and the negative charge on planet’s surface. During cloudless skies days, the air possesses a voltage of nearly around 100 volts for each and every meter from above the ground.

Ballooning spiders process within this planetary electric field. When their silk relieve their bodies then it picks up a negative charge. This oppose the similar negative charges on the surfaces on which the spiders settles and create sufficient force to lift them into the air. And spiders can hike those forces by climbing onto blades of grass,twigs, or leaves.

6 0

3 years ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

2 years ago

Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120

Dimas [21]

Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) $\left | a\right |=5\ km$

the magnitude of the second vector(say) $\left | b\right |=7\ km$

the angle between them is $120^{\circ}$

The resultant vector magnitude is given by

$\left | \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}$

$\left | \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left | \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left | \vec{R}\right |=6.24\ km$

4 0

2 years ago