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cluponka [151]
2 years ago
5

Which of the following provides evidence that there must be at least two types of electrical charge, but that there is only one

type of mass?
A.) The magnitude of the Coulomb's law constant differs greatly from the magnitude of the universal gravitational constant.
B.) Contact forces such as friction are caused primarily by electrostatic forces and not by gravitational ones.
C.) The electric force has a relative strength that is much greater than the strength of the gravitational force.
D.)On the scale of the solar system, electric forces cancel out and gravitational forces dominate.
Physics
1 answer:
polet [3.4K]2 years ago
5 0

Answer:

Option D (On the...............dominate) would be the right approach.

Explanation:

The Gravitational constant (G) will be:

= 6.67\times 10^{-11}

The Coulomb's law constant (K) will be:

= 9\times 10^9

  • Throughout particular, these have been determined that among 2 substances with almost the similar form of charge, the combination of electromagnetic as well as the force does seem to be usually the following:

⇒ \frac{f_e}{f_g}\sim 10^{42}

  • By that same argument, the electrostatic force including its planet's atmosphere would have strongly influenced the effect, as well as maybe the planet's atmosphere, would have crashed, or perhaps the earth would have shifted at a much longer exposure from one another and.
  • Throughout particular, astronomical distance statutory framework that gravity seems to be predominant, whereas electrostatic forces have been generally ignored. It is quite since there are so many categories of allegations throughout the planet's atmosphere that balance out someone else's effects, there's only yet another form of momentum, because although the forces are still cumulative, as well as therefore offering to help everything hold to the universe, encouraging the universe just to rotate across the sun.

The latter three choices aren't connected to either the situation mentioned in the clarification segment elsewhere here.

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The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

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On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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