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zaharov [31]
3 years ago
5

A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

of the electric field at the center of the circle?

Physics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

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Explanation:

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When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the s
skad [1K]

Answer:

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Explanation:

From the question we are told that

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Generally the net force acting on the sky diver is mathematically represented as

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If Two stars, Betty and Wilma, are both on the main sequence. Betty is more luminous than Wilma. Which one has a hotter surface
monitta

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Betty

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Since they're both on the most sequence but Betty is more luminous than Wilma, Betty must be located to a higher place on the most sequence. Therefore, Betty encompasses a hotter surface temperature, is more massive, and encompasses a larger radius. Betty also will evolve faster than Wilma, and if they were formed at the identical time Betty will put off the most sequence first

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A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at
Olegator [25]

Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to 1 × 10^{-12} meter

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To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by 174 × 10^{-12} meter we get that the number of sheets will be as follows

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