Question

two charges +6 micro coulomb and -2 micro coulomb are seperated by a distance of 30cm apart in air. find the electrostatic force between the charges?​

Answer 1

The electrostatic force between the charges is 3.6N

Explanation:

We know that electrostatic force between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by

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

Where k is coulomb's constant = $9\times 10^9$ Nm²/C²

Thus, the force between the charges 6 μC and -2 μC, separated by a distance 30 cm = 0.3 m is given by

$F=9\times 10^9\times\frac{6\times 10^{-6}\times 2\times 10^{-6}}{(3\times 10^{-1})^2}$

$\implies F=\frac{9\times 6\times 6\times 10^{-3}}{3\times 3\times 10^{-2}}$ N

$\implies F=3.6$ N

Since both the charges are of opposite sign

Therefore, this force will be attractive.

Hope this answer is helpful.

Know More:

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Answer 2

Answer:

1.2N

Explanation:

F =KqQ/r2

F=9x10^9X6x10^-6X-2x10^-6 / [30x10^-2]^2

F =-108X10 ^-3 / 900x10^-4

F=-1080 / 900

F=1.2 N