Question

8.The magnitude of charge which exerts a force of 1/9 N on a charge of 100 µC when kept in air a distance of 9 cm from it is ----------

Answer 1

10cm

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Answer 2

The magnitude of another charge is $10^{-9}\ C$

Explanation:

Given that,

Force between two charges, $F=\dfrac{1}{9}\ N$

First charged particle, $q_1=100\ \mu C=100\times 10^{-6}\ C=10^{-4}\ C$

Let $q_2$ is another charged particle

Distance between charges, d = 9 cm = 0.09 m

If F is the electrostatic force between charged particles. It is given by :

$F=\dfrac{kq_1q_2}{d^2}$

$q_2=\dfrac{Fd^2}{kq_1}$

$q_2=\dfrac{\dfrac{1}{9}\times (0.09)^2}{9\times 10^9\times 10^{-4}}$

$q_2=10^{-9}\ C$

So, the magnitude of another charge is $10^{-9}\ C$. Hence, this is the required solution.

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Electrostatic force

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