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oksano4ka [1.4K]
3 years ago
5

According to Coulomb’s Law, what happens to the force when the distance increase between 2 particles?

Physics
1 answer:
ohaa [14]3 years ago
8 0

Answer:

The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, <u>the attraction or repulsion becomes weaker</u>, decreasing to one-fourth of the original value.

Explanation:

Coulomb’s law, mathematical description of the electric force between charged objects. Formulated by the 18th-century French physicist Charles-Augustin de Coulomb, it is analogous to Isaac Newton’s law of gravity.

Both gravitational and electric forces decrease with the square of the distance between the objects, and both forces act along a line between them. In Coulomb’s law, however, the magnitude and sign of the electric force are determined by the electric charge, rather than the mass, of an object. Thus, charge determines how electromagnetism influences the motion of charged objects. Charge is a basic property of matter. Every constituent of matter has an electric charge with a value that can be positive, negative, or zero.

Coulomb's Law says that the force between 2 charges is proportional to the product of the quantities of charge on each and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is F=k\frac{q_{1}q_{2}   }{r^{2} }.

F is the force.

k is the Coulomb's constant (8.987*10^{9} \frac{Nm^{2} }{C^{2} }).

q_{1} is the electric charge of object 1.

q_{2} is the electric charge of object 2.

r is the distance between the two charges.

Electric force is inversely proportional to (r^{2}) instead of (r). As the distance between charges increases, the electric force decreases by a factor of \frac{1}{r^{2} }.

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An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi
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Answer:

a) \vec F_{B} = 8.766\times 10^{-14}\,T\,k, b) \vec F_{B} = -8.766\times 10^{-14}\,T\,k

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

\vec F_{B} = q\cdot \vec v \times \vec B

The charge of the electron is equal to -1.602\times 10^{-19}\,C. Then, cross product can be solved by using determinants:

\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = 8.766\times 10^{-14}\,T\,k

b) The charge of the proton is equal to 1.602\times 10^{-19}\,C. Then, cross product has the following determinant:

\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = -8.766\times 10^{-14}\,T\,k

8 0
3 years ago
in a young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance th
blagie [28]

Answer:

Zero

Explanation:

Because using

Deta X= dsinစ x n(lambda)

But we know that for central maxima

n is zero

So after substituting

Deta x = 0

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3 years ago
Which statement accurately describes the motion of the object in the graph above over 10 seconds? Group of answer choices The ob
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Answer:

Object appears to move forward at 1 cm/sec, then the velocity drops to zero for 3 sec and then moves forward at 2 cm/sec     (11 - 3) / (10 - 6) = 2 cm/sec

7 0
3 years ago
a large truck has a mass of 20000kg. it is traveling at 28m/s along a staright road . calculate the kinetic energy ​
Semmy [17]

Answer:

The answer is

<h2>7,840,000 J</h2>

Explanation:

The kinetic energy KE of an object given it's mass and velocity can be found by using the formula

KE =  \frac{1}{2} m {v}^{2}  \\

where

m is the mass

v is the velocity

From the question

m = 20000kg

v = 28 m/s

It's kinetic energy is

KE =  \frac{1}{2}  \times 20000 \times  {28}^{2}  \\  = 10000 \times 784

We have the final answer as

<h3>7,840,000 J</h3>

Hope this helps you

3 0
3 years ago
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