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oksano4ka [1.4K]
2 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]2 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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The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

\displaystyle V=\frac{kq}{r}

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

V_2=-22090.91\ V

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

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V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V

V=-11045.45 \ V

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An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and
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Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

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= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

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Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

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magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

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To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

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Taking the relative initial velocity of the man and the woman as 0 m/s, we have;

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The final velocity of the man = v₂₁

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Which gives;

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65 × 0 + 45 × 0 = 65 × v₂₁ + 45 × 1.5

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The final velocity of the man, her partner = v₂₁ ≈ -1.04 m/s.

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