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Two point charges, initially 2 cm apart, are moved to a distance of 10 cm apart. By what factor does the resulting electric force
between them change?
a 25
b.5
C.1/5
D.1/25

Respuesta :

abidemiokin

Answer:

1/25

Explanation:

Coulombs law states that the force of attraction between two charges q and Q is directly proportional to the product of the charges and inversely proportional to the square of the distance r between them.

It is expressed as;.

F = kqQ/r²

If two point charges, initially 2 cm apart, the force between them will be;

F = kqQ/2²

F = kqQ/4 ... (1)

If the charges are now moved to a distance of 10cm apart, the force between them will then become;

F2 = kqQ/10²

F2 = kqQ/100.. (2)

Dividing equation 2 by 1;

F2/F = (kqQ/4)/kqQ/10)

F2/F = kqQ/100 × 4/kqQ

F2/F = 1/25

F2 = (1/25)F

The resulting electric force

between the changes by 1/25 since that serves as the constant of proportionality between the two forces

The factor does the resulting electric force between them change is 1/25

Calculation of factor:

Here we apply the Coulombs that defined  the force of attraction between two charges q and Q i.e.  directly proportional to the product of the charges and inversely proportional to the square of the distance.

Now

It is expressed as;.

F = kqQ/r²

In the case when two point charges, initially 2 cm apart, the force between them should be

F = kqQ/2²

F = kqQ/4 ... (1)

Now in the case when the charges are now moved to a distance of 10cm apart, so the force between them should be

F2 = kqQ/10²

F2 = kqQ/100.. (2)

Now here we Dividing equation 2 by 1;

So,

F2/F = (kqQ/4)/kqQ/10)

F2/F = kqQ/100 × 4/kqQ

F2/F = 1/25

F2 = (1/25)F

Learn more about the force here: https://brainly.com/question/22614246

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