Physics, asked by rajandhanya93, 10 months ago

3 point charges +2 micro coulomb each placed at the corners of equilateral triangle of side 1 m .Find the magnitude of force between charges q1 and q2 and q1 and q3?

Answers

Answered by BrainlyConqueror0901
9

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

\green{\tt{\therefore{F_{1}=F_{2}=0.36\:N}}}

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

 \green{\underline \bold{Given :}} \\  \tt:\implies Point \: charges =  + 2  \: \mu \:  C\\  \\ \tt: \implies distance \: between \:charges = 1 \: m   \\  \\  \red{\underline \bold{To \: Find:}} \\  \tt:  \implies   F_{1}   =  ?\\\\  \tt:  \implies   F_{2}   =  ?

• According to given question :

 \bold{As \: we \: know \: that} \\  \tt: \implies  F_{1}   =  \frac{k q_{1} q_{2}}{ {r}^{2} }    \\  \\ \tt: \implies F_{1} = \frac{k \times 2 \times 2}{ {1}^{2} }  \\  \\ \tt: \implies  F_{1} = 4k\\\\ \tt:\implies F_{1}=4\times 10^{-12}\times 9\times 10^{9}\\\\  \green{\tt:\implies F_{1}=0.36\:N}\\  \\  \bold{Similarly:} \\  \tt: \implies  F_{2}   =  \frac{k q_{1} q_{2}}{ {r}^{2} }    \\  \\ \tt: \implies F_{2} = \frac{k \times 2 \times 2}{ {1}^{2} }  \\  \\ \tt: \implies  F_{2} = 4k \\  \\  \green{\tt: \implies  F_{2}= 0.36\:N}\\\\\bold{For \: magnitude \: of \: fores} \\  \tt : \implies  F_{net} =  2F_{1}  \cos \:  \frac{60}{2}  \\  \\ \tt : \implies  F_{net} = 2 \times 4k \times  \frac{ \sqrt{3} }{2}  \\  \\ \tt : \implies  F_{net} = 4 \sqrt{3}  k \\  \\ \tt : \implies  F_{net} =  4 \times  {10}^{ - 12}  \times  \sqrt{3}  \times 9 \times  {10}^{ 9}  \\  \\  \tt : \implies  F_{net} =  36  \sqrt{3} \times  {10}^{ - 3}  \\  \\  \green{\tt : \implies  F_{net} = 3.6 \sqrt{3}  \times  {10}^{ - 1} \: N}

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