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2 years ago

True/False

Physics

1 answer:

Alborosie2 years ago

6 0

Answer:

True

Explanation:

Scientists and professionals in health sciences are invited to participate in congresses, conferences and events. Communications to conferences can be of different types, but it is always necessary to draw a table with the contents that you want to present.

Due to the rise and importance of conferences in recent years, scientists and scientific societies that provoke conferences have gained greater knowledge of the importance of defining their publications, so that the problems related to these issues are becoming smaller .

It is important to know how to write the communication and if the conference proceedings or reports will be considered as primary publication, then it is recommended to prepare the manuscript in the same style as the journal articles. For this it is recommended to use the same method.

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Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

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Explanation:

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Answer Explanation :

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Q = $\frac{(P_2-P_1) r^{4}\times \pi}{8\times\eta \times l }$

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