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kozerog [31]
3 years ago
9

An electroscope is a simple device consisting of a metal ball that is attached by a conductor to two thin leaves of metal foil p

rotected from air disturbances in a jar, as shown. When the ball is touched by a charged body, the leaves that normally hang straight down spread apart. Why? (Electroscopes ate useful not only as charge detectors but also for measuring the quantity of charge: the more charge transferred to the ball, the more the leaves diverge.)
Physics
1 answer:
baherus [9]3 years ago
3 0

Answer:

the electroscope separate  by the presence of charge carriers

Explanation:

Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where

                  Fe - Tx = 0

                  Fe = Tx

In summary, the electroscope separate its leaves by the presence of charge carriers

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Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
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Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

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