PHYSICS HIGH SCHOOL

Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are separated by 12 cm. (Why can you solve this problem without knowing the magnitudes of the charges

Answers

Answer 1
Answer:

The force between the two point charges is "5 N".

Force and magnitudes:

The total of almost all of the forces operating on an item is referred to as the magnitude of force. When some of the forces operate in the same direction, this same magnitude rises. Whenever forces occur in opposite directions around an item, the amount of something like the force diminishes.

Distance between charges, d = 6 cm

Attractive force, F = 20 N

Separated, d' = 12 cm

We know the formula,

F = k (q_1q_2)/(d^2)

or,

F \propto (1)/(d^2) ...(Equation 1)

New force will be:

F' \propto (1)/(d'^2) ...(Equation 2)

From "Equation 1" and "Equation 2", we get

→ (F)/(F') =(d'^2)/(d^2)

By substituting the values,

  (20)/(F') =(12^2)/(6^2)

  (20)/(F') =(144)/(36)

  F' = 5 N

Thus the approach above is appropriate.

Find out more information about force here:

brainly.com/question/388851
Answer 2
Answer:

Answer:

The force between them when they are separated by 12 cm is 5 N.

Explanation:

Distance between two point charges, d = 6 cm

The attractive force between them is 20 N, F = 20 N

Let F' is the force between them when they are separated by 12 cm, d' = 12 cm

The force between point charges is given by the formula as :

F=k(q_1q_2)/(d^2)

F\propto (1)/(d^2)........(1)

New force,

F'\propto (1)/(d'^2)............(2)

From (1) and (2) :

(F)/(F')=(d'^2)/(d^2)\n\n(20)/(F')=(12^2)/(6^2)\n\nF'=5\ N

So, the force between them when they are separated by 12 cm is 5 N. Hence, this is the required solution.

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Explanation:

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Answers

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Answers

Answer:

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