Two equal charges are placed in air are separated by a distance of 3 meter ,repel each other with a force of $0.19 f$ .calculate the magnitude of either of the charge ( $9.9 \times 10^{- 1} C)$

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Solution

Given, $d = 3 m, f = 0.19 f$

So, $f = \frac{k q^{2}}{r^{2}} \Rightarrow k q^{2} = 0.19 \times 9 \Rightarrow q^{2} = \frac{1.71}{9.0 \times 10^{9}}$ Where $k = 9.0 \times 10^{9}$

$\Rightarrow q = \sqrt{0.19 \times 10^{- 9}}$ C

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Two equal charges are placed in air are separated by a distance of 3 meter ,repel each other with a force of 0.19f.calculate the magnitude of either of the charge (9.9×10−1C)

Given, d=3m,f=0.19fSo, f=kq2r2⇒kq2=0.19×9⇒q2=1.719.0×109 Where k=9.0×109⇒q=√0.19×10−9C

Two equal charges are placed in air are separated by a distance of 3 meter ,repel each other with a force of $0.19 f$ .calculate the magnitude of either of the charge ( $9.9 \times 10^{- 1} C)$

Given, $d = 3 m, f = 0.19 f$

So, $f = \frac{k q^{2}}{r^{2}} \Rightarrow k q^{2} = 0.19 \times 9 \Rightarrow q^{2} = \frac{1.71}{9.0 \times 10^{9}}$ Where $k = 9.0 \times 10^{9}$

$\Rightarrow q = \sqrt{0.19 \times 10^{- 9}}$ C