Question

The electric field line near two equal and opposite charges placed in air at a distance 5cm apart. Calculate the magnitude of the charges if the force between them is 90N

Answer 1

Explanation:

In this way we can solve your problem my friend....

Answer 2

Given ,

The two equal and opposite charges are separated by distance 5 cm and the force between them is 90 N

We know that , The force between two charges is given by

$\fbox{F = \frac{k q_{1} q_{2} }{ {(r)}^{2} } }$

Thus ,

$\sf \mapsto 90 = \frac{9 \times {(10)}^{9} \times {(q)}^{2} }{ { \{ 5 \times {(10)}^{ - 2}\}}^{2} } \\ \\\sf \mapsto 25 = {(10)}^{12} \times {(q)}^{2} \\ \\\sf \mapsto q = \sqrt{25 \times {(10)}^{-12} } \\ \\ \sf \mapsto q = 25 \times {(10)}^{-6} \: \: c$

$\therefore \sf \underline{The \: magnitude \: of \: charge \: is \: 5 \times {(10)}^{-6} \: c }$