Question

Two point charges 2uC and 6uC and seperated by a distance of 10 cm then find force on charge 2uC due to 6uC. ​

Answer 1

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

$\green{\tt{\therefore{F_{26}=10.8\:N}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Two \: point \: charge = 2 \mu \: c \: and \: 6 \mu \: c \\ \\ \tt: \implies Distance \: between \: charges(r) = 10 \: cm \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Force \: due \: to \: 6 \mu \: C \: on \: 2 \mu \: C( F_{26}) =?$

• According to given question :

$\tt \circ \: r = \frac{10}{100} = 0.1 \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies F_{26} = \frac{k q_{1} q_{2} }{ {r}^{2} } \\ \\ \tt: \implies F_{26} = \frac{k \times 2 \times {10}^{ - 6} \times 6 \times {10}^{ - 6} }{ {(0.1)}^{2} } \\ \\ \tt: \implies F_{26} = \frac{k \times 12 \times {10}^{ - 12} }{0.01} \\ \\ \tt \circ \: k = 9 \times {10}^{9} \\ \\ \tt: \implies F_{26} = \frac{9 \times {10}^{9} \times 12 \times { 10 }^{ - 12} }{1 \times {10}^{ - 2} } \\ \\ \tt: \implies F_{26} =108 \times {10}^{9 - 12 + 2} \\ \\ \tt: \implies F_{26} = 108 \times {10}^{ - 1} \\ \\ \green{\tt: \implies F_{26} =10.8 \: N} \\ \\ \green{\tt \therefore Force \: on \: charge \: 2 \mu \: c \: due \: to \: 6 \mu \: c \: is \: 10.8 \: N}$