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College Physics (Urone)
Found in: Page 664

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Short Answer

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?

If the separation is increased to three times of the original, force will become 0.556N .

See the step by step solution

Step by Step Solution

Step 1: Given Data

  • Value of point charges is 5.00 N

Step 2: Electrostatic force

Coulomb stated that when two-point charges are separated by some distance, they experience some force of attraction or repulsion. This force of attraction or repulsion is known as electrostatic force.

Step 3: New force

Initially, when the charges q and Q are separated by some distance r the electrostatic force between them is,


Here, F is the electrostatic force (F = 5.00 N) , and K is the electrostatic force constant.

When the separation between the charges increased by a factor of three r'=3r. The new electrostatic force (F') between the point charges is,

F'=KqQr'2 =KqQ3r2 =KqQ9r2

Using equation (1.1),


Substituting 5.00 N for F,

F'=5.00 N9 =0.556 N

Hence, when the separation becomes three times of the original, the new force will be 0.556 N.

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