Question

Two equal charges are placed at a distance of 3m. A repulsive force of 1.6N acts on it. The value of each charge will be: (a) 2C (b) 4C (c) 40C (d) 80C​

Answer 1

Answer:

$f ={ kq }^{2} \div {r }^{2} \\1.6 = 9 \times {10}^{9} \times {q }^{2} \div 9 \\ ie \: \: {q}^{2} = 1.6 \times {10}^{ - 9 } \\ q = 0.4 \times {10}^{ - 4 } \\ hope \: it \: helps \: u.... \:</p><p> pls \: mark \: it \: brainliest....$☺☺☺☺☺

Answer 2

The value of each charge will be $4\times 10^{-5}\ C$.

Explanation:

Distance between two charges, r = 3 m

Repulsive force between charges, F = 1.6 N

We need to find value of each charge. The electric force between two charges is given by :

$F=\dfrac{kq^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{1.6\times (3)^2}{9\times 10^9}}$

$q=4\times 10^{-5}\ C$

So, the value of each charge will be $4\times 10^{-5}\ C$. Hence, this is the required solution.

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