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gavmur [86]
3 years ago
6

How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their or

iginal distance of separation? 1. Reduces to one quarter of original value 2. Doubles 3. Doesn’t change 4. Reduces to one half of original value 5. Quadruple
Physics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

5. Quadruple

Explanation:

The electrostatic force between two charged particles is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

If the distance between the charges is reduced to half,

r' = \frac{r}{2}

So the new force will be

F'=k\frac{q_1 q_2}{(r/2)^2}=4(k\frac{q_1 q_2}{r^2})=4F

So, the force will quadruple.

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Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

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we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

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      d₂ = M /m d₁

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