Question

Electric charges 1mc,-1mc and 2mc are in air at corners a,band c of equilateral triangle abc having length 10 cm resultant force o charge at c is

Answer 1

★$\color{darkblue}\underline{\underline{\sf Given-}}$

★$\underline{\sf In \: equilateral \: traingle}$

• Charge on point A = 1mC
• Charge on point B = -1mC
• Charge on point C = 2mC
• Lenght of equilateral traingle = 10cm

★$\color{darkblue}\underline{\underline{\sf To \: Find -}}$

• Resultant force at point C

★$\color{darkblue}\underline{\underline{\sf Formula \: Used -}}$

Coulomb's Law

$\color{violet}\bullet\underline{\boxed{\sf F=9×10^9\dfrac{Q_1Q_2}{r^2}}}$

★$\color{darkblue}\underline{\underline{\sf Solution-}}$

★$\implies{\sf F_A=Force \: on \: point \: C \: due \; to \: Charge \: A}$

$\implies{\sf F_A=9×10^9\dfrac{10^{-3}×2×10^{-3}}{(10×10^{-2})^2}}$

$\implies{\sf \dfrac{18×10^3}{100×10^{-4}}}$

$\implies{\sf \dfrac{18×10^3}{10^{-2}}}$

$\color{orange}\implies{\sf F_A= 18×10^5\:N }$

★$\implies{\sf F_B=Force \: on \: point \: C \: due \; to \: Charge \: B}$

$\implies{\sf F_B=9×10^9\dfrac{10^{-3}×2×10^{-3}}{(10×10^{-2})^2}}$

$\implies{\sf \dfrac{18×10^3}{100×10^{-4}}}$

$\implies{\sf \dfrac{18×10^3}{10^{-2}}}$

$\color{orange}\implies{\sf F_B=18×10^5\:N }$

➩ Net force on point C

$\implies{\sf F_{net}=\sqrt{(F_A)^2+(F_B)^2+2F_1F_2cos120°}}$

$\implies{\sf F_{net}=\sqrt{(18×10^5)^2+(18×10^5)^2+2×18×10^5×cos120°}}$

cos120° = -½

$\implies{\sf F_{net}=\sqrt{(18×10^5)^2+(18×10^5)^2+2×18×10^5×\left(-\dfrac{1}{2}\right)}}$

$\implies{\sf F_{net}=\sqrt{(18×10^5)^2+(18×10^5)^2-(18×10^5)^2}}$

$\implies{\sf \sqrt{(18×10^5)^2 }}$

$\color{red}\implies{\sf F_{net}=18×10^5\:N}$

★$\color{darkblue}\underline{\underline{\sf Answer-}}$

★Net force on charge C is $\color{red}{\sf 18×10^5\:N}$

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

Given ,

• The three point charges i.e 1 mc ,-1 mc and 2 mc are located at the corners of equilateral triangle ΔABC

• The length of side of equilateral triangle is 10 cm

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

$\large \mathtt{ \fbox{F = k \frac{ q_{1}q_{2}}{ {(r)}^{2} } }}$

Thus ,

The force on C due to A is

$\sf \hookrightarrow F_{CA} = \frac{9 \times {(10)}^{9} \times 1 \times {(10)}^{ - 3} \times 2 \times {(10)}^{ - 3} }{ {(10 \times {(10)}^{ - 3} )}^{2} } \\ \\ \sf \hookrightarrow F_{CA} = \frac{18 \times {(10)}^{3} }{ {(10)}^{ - 2} } \\ \\ \sf \hookrightarrow F_{CA} =18 \times {(10)}^{5} \: \: newton$

Since , the magnitude of charge at A is same as charge at point B

Thus ,

$\sf \underline{ \fbox{F_{CA} = F_{CB} =F \: ( Let) }}$

The resultant force of $\mathtt{ F_{CA} }$ and $\mathtt{ F_{CB} }$ is

$\sf \hookrightarrow Force= \sqrt{ {(F)}^{2} + {(F)}^{2} + 2 {(F)}^{2} \cos(120) } \\ \\ \sf \hookrightarrow Force= \sqrt{2 {(F)}^{2} - \cancel 2 {(F)}^{2} \times \frac{1}{ \cancel2} } \\ \\ \sf \hookrightarrow Force= \sqrt{ {(F)}^{2} } \\ \\ \sf \hookrightarrow Force=F \\ \\ \sf \hookrightarrow Force= 18 \times {(10)}^{5} \: \: newton$

Hence , the force on C due to A and B is 18 × (10)^5 Newton

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