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valina [46]
2 years ago
8

When you rub neutral objects​

Physics
1 answer:
FrozenT [24]2 years ago
4 0

electrons can be knocked loose from one object and picked up by the other. The object that gains electrons becomes negatively charged, while the object that loses electrons becomes positively charged.

b gained electrons

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Two particles having charges of 0.50~\text{nC}0.50 nC (q_1q ​1 ​​ ) and 10~\text{nC}10 nC (q_2q ​2 ​​ ) are separated by a dista
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Answer:

The third charge must be placed 0.548 m from q₁.

Explanation:

Let r = 3m  be the distance between charge q₁ and q₂.

Let x be the distance between charge q₁ and charge q₃ (the third positive charge)

Then r - x is the distance between charge q₂ and q₃

Let the electrostatic force between q₁ and q₃ be F = kq₁q₃/x²

Let the electrostatic force between q₂ and q₃ be F' = kq₂q₃/(r - x)²

Since F + (-F') = 0 (the signs on the forces are different since the forces are in opposite directions)which is required when the net force on q₃ is zero, then

F - F' = 0

F = F'

So, kq₁q₃/x² = kq₂q₃/(r - x)²

q₁/x² = q₂/(r - x)²

[(r - x)/x]² = q₂/q₁

taking square-root of both sides,

(r - x)/x = ±√q₂/q₁

r/x - 1 = ±√q₂/q₁

r/x = 1 ±√q₂/q₁

x = r/(1 ±√q₂/q₁)

substituting the values of the variables r = 3 m, q₁ = 0.50 nC and q₂ = 10 nC

x = 3 m/(1 ±√10 nC/0.5 nC)

x = 3 m/(1 ±√20)

x = 3 m/(1 ± 4.472)

x = 3 m/5.472 or 3 m/-3.472

x = 0.548 m or -0.864 m

So the third charge must be placed 0.548 m to the right of q₁ or 0.864 m to the left of q₁.

Since we are concerned about the line of charge that connects q₁ and q₂, the third charge must be placed 0.548 m from q₁.

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