Question

Find force between two charges 3microcoulomb and 2 coulomb at 7cm.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{F=1.1\times 10^{7}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies Two \: charges = 3 \mu C\: and \: 2 \: C \\ \\ \tt: \implies Distance(r) = 7 \: cm \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies Force \: between \: charges(F) =?$

• According to given question :

$\tt \circ \: q_{1} = 3 \mu C = 3 \times {10}^{ - 6} \:C\\ \\ \tt: \circ \: q_{2} = 2 \: C \\ \\ \tt \circ \: r = \frac{7}{100} = 0.07 \: m\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies F = \frac{ kq_{1} q_{2} }{ {r}^{2} } \\ \\ \bold{ Where : } \\ \tt \circ \: k = 9 \times {10}^{9} \\ \\ \tt \circ \: F \propto \frac{1}{ {r}^{2} } \\ \\ \tt: \implies F = \frac{ 9 \times {10}^{9} \times 3 \times {10}^{ - 6} \times 2}{ {(0.07)}^{2} } \\ \\ \tt: \implies F= \frac{9 \times 3 \times 2 \times {10}^{9 - 6} }{0.0049} \\ \\ \tt: \implies F = \frac{54 \times {10}^{3} }{49 \times {10}^{ - 4} } \\ \\ \tt: \implies F= \frac{54}{49} \times {10}^{3 + 4} \\ \\ \green{\tt: \implies F = 1.1 \times {10}^{7} \: N} \\ \\ \green{\tt \therefore Force \: between \: charges \: is \: 1.1 { \times 10}^{7} \: N}$