Question

two charged bodies q1 and q2 are separated by a distance of 2m experience a force of repulsion equal to 1.8 n what will be the force of repulsion be when they are 10 m apart​

Answer 1

F₁ = ( k q₁ q₂ ) ÷ (r₁)²

1.8 N = (k q₁ q₂ ) ÷ 2²

1.8 × 4 = (k q₁ q₂ )

7.2 = (k q₁ q₂ )  

F₂ = ( k q₁ q₂ ) ÷ (r₂)

    = 7.2 ÷ 10²

    = 7.2 ÷ 100

    = 0.072 or 7.2 × 10⁻²

Hope this helps!

Answer 2

Answer:

 $7.2 * 10^-2 N.$ is the force of repulsion.

Explanation:

The force between two charged particles or bodies  is given by Coulomb's law which states that:

$F = k * q1 * q2 / r^2$

where F is the force, k is the Coulomb constant $(9 * 10^9 N m^2/C^2)$, q1 and q2 are the charges of the particles, and r is the distance between them.

Given that q1 and q2 are separated by 2m and experience a force of 1.8 N, we can use Coulomb's law to find their charges:

$1.8 N = k * q1 * q2 / 2^2$

Solving for q1 * q2, we get:

$q1 * q2 = (1.8 N * 4) / k$

$q1 * q2 = 8.0 * 10^-7 C^2$

Now, if the distance between q1 and q2 is increased to 10m, the force of repulsion between them can be calculated using Coulomb's law:

$F' = k * q1 * q2 / r'^2$

where r' is the new distance (10m).

Substituting the values we have, we get:

$F' = (9 * 10^9 N m^2/C^2) * (q1 * q2) / (10m)^2$

$F' = (9 * 10^9 N m^2/C^2) * (8.0 * 10^-7 C^2) / 100m^2$

$F' = 7.2 * 10^-2 N$

Therefore, the force of repulsion between q1 and q2 when they are 10m apart is $7.2 * 10^-2 N.$

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