Two point charges, initially 2.0 cm apart, experience a 1.0-n force. if they are moved to a new separation of 8.0 cm, what is the electric force between them?

Respuesta :

The electric force between two point charges is 1/6 N.

Given data:

Initial distance between the charges (r1) = 2.0 cm

The new distance between the charges (r2) = 8.0 cm

Initial electric force between the charges - F1 = 1.0 N

Lets' say the two charges are q and q2. The formula to calculate electric force between the two charges is given below

F = K |q1| |q2| / r^2

K is a constant.

New  electric force can be expressed as

F2/F1 = K |q1| |q2| / r2^2 ÷  K |q1| |q2| / r1^2

F2/F1 = r2^2 / r1^2

After substituting the values we will get

F2 = 1.0 N ( 2.0 / 8.0)^2

F2 = 4/64 N

F2 = 1/16N

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