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A positive charge of 6.0 × 10-4 C is in an electric field that exerts a force of 4.5 × 10-4 N on it. What is the strength of the

VLD [36.1K]

Answer:

0.75 NC⁻¹

Explanation:

Electric field intensity ( or strength of the electric field ) is the force per a 1 C charge,

So, Force (F) = Electric field intensity(E) × Charge (q)

F = E×q ⇒ q = F/E

= 4.5×10⁻⁴/6×10⁻⁴ = 0.75 NC⁻¹

According to cool om's law electric fields are generated due to charges. When charges are same there is a repulsive force acted on both charges. When charges are opposite there is a attraction force acted on both charges.

According to cool om's law,

F =G×q1×q2 / r²

F = force exerted of two charges

q1 , q2 = charges

r = distance between two charges

And also Electric field intensity is a vector which has a magnitude and direction both. Direction is depending on a charge and the sign of the charge

6 0

3 years ago

Read 2 more answers

What are the three reasons cells divide?

hoa [83]

1 growth. Go from one cell/( zygote to a trillion)
2 replace. Repair\ 50 million cells die second.
3 reproduction. ( make cells for reproduction make specialized sex cells)

7 0

4 years ago

For an object to appear red, what must happen?

Lapatulllka [165]

Answer:

<h2>C. It must reflect red light wavel</h2>

Explanation:

The color of the object is reflected by the object other colors are absorbed by the object.

Here object appears in red color so red color is reflected by the object and other colors are absorbed.

3 0

3 years ago

Mercury has an average disease to the sun of 0.39 AU. In two or more complete sentences, explain how to calculate the orbital pe

alukav5142 [94]

Answer: 88 Earth days

Explanation:

According to the Kepler Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit:

$T^{2}=a^{3}$ (1)

If we assume the orbit is circular and apply Newton's law of motion and the Universal Law of Gravity we have:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (2)

Where $M$ is the mass of the massive object and $G$ is the universal gravitation constant. If we assume $M$ constant and larger enough to consider $G$ really small, we can write a general form of this law:

$MT^{2}=a^{3}$ (3)

Where $T$ is in units of Earth years, $a$ is in AU (1 Astronomical Unit is the average distane between the Earth and the Sun) and $M$ is the mass of the central object in units of the mass of the Sun.

This means when we are making calculations with planets in our solar system $M=1$ .