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3 years ago

A +0.0129 C charge feels a 4110 N

Physics

1 answer:

Savatey [412]3 years ago

4 0

Answer:

r = 14.13 m

Explanation:

Given that,

Charge 1, q₁ = +0.0129 C

Charge 2, q₂ = -0.00707 C

The force between charges, F = 4110 N

We need to find the distance between charges. The formula for the force between charges is given by :

$F=k\dfrac{q_1q_2}{r^2}$

Where

r is the distance between charges

So,

$r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 0.0129 \times 0.00707 }{4110 }} \\\\r=14.13\ m$

So, the distance between charges is equal to 14.13 m.

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