Question

6. Two equal and similar charges 0.03m apart in air, repel each other with a force of 45kgf. Find
charge in coulomb.​

Answer 1

$\huge \fcolorbox{red}{pink}{Solution :)}$

Given ,

The distance between two charged particles is 0.03 m and the force is 45 Kgf or 441 Newton

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

$\large \mathbb{ \fbox{F = \frac{k( Q_{1} \times Q_{2} )}{ {(r)}^{2} } } }$

Let us assume that , the magnitude of two equal charges be Q

Now , substitute the known values

$\sf \hookrightarrow 441 = \frac{9 \times {(10)}^{9} \times {(Q )}^{2} }{ {(0.03)}^{2} } \\ \\ \sf \hookrightarrow 441 = \frac{9 \times {(10)}^{9} \times {(10)}^{4} \times {(Q )}^{2} }{9} \\ \\ \sf \hookrightarrow 441 = {(10)}^{13} \times {(Q )}^{2} \\ \\ \sf \hookrightarrow (Q) = \sqrt{ \frac{441}{ {(10)}^{13} } } \\ \\ \sf \hookrightarrow (Q) = \frac{21 }{3162277.66} \\ \\ \sf \hookrightarrow (Q) = 6.640783 \times {(10)}^{ - 6} \: \: coulomb$

Hence , the magnitude of two equal charges is 6.640783 × (10)^(-6)

Answer 2

Answer:

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