Protons and neutrons both have positive charge, and electrons have negative charge. true or false

Physics

2 answers:

Mice21 [21]2 years ago

8 0

The given statement "protons and neutrons both have positive charge, and electrons have negative charge" is false.

Answer: Option 2

<u>Explanation:</u>

The atoms of all elements are made up of tiny particles that were unknown to the world for many years. With the efforts of wise brains such as J.J. Thomson, Rutherford and Chadwick, it has been concluded that an atom is not the smallest particle and is comprised of three tiny particles’ combination- Protons, Electrons and Neutrons.

The protons have a natural positive charge that attracts the equal but opposite charge towards them and they are electrons. Electrons have a negative charge and revolve around the nucleus of an atom in their constituent orbits.

Neutrons don’t have any charge on them and hence, remain at rest inside the nucleus of atoms. They are actually the flipped iso-spins of protons. The purpose of neutrons in the nucleus is to create a balance between two or more protons so that they can co-exist in the nucleus and it may not scatter.

Kruka [31]2 years ago

7 0

Answer:

False

Explanation:

Protons are positively charged, electrons are negatively charged, and neutrons have no charge.

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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

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;

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