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3 years ago

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The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of

dielectric constant 9, what will be the value of new coulomb force?

1 answer:

vodomira [7]3 years ago

3 0

Answer:

The new Coulomb force is q₁q₂/9πε₀r²

Explanation

The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².

Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.

Since we have air between q₂ and q₃, the coulomb force between them is

F' = q₂q₃/4πε₀d²

= q₂(q₁/κ)/4πε₀(r/2)²

= 4q₂q₁/κ4πε₀r²

= 4/κ(q₂q₁/4πε₀r²)

= 4/9 × (q₂q₁/4πε₀r²)

= q₁q₂/9πε₀r²

So, the new Coulomb force is q₁q₂/9πε₀r²

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Explanation:

Have a look to the diagrams attached below.

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Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
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for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

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c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

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The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

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The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago