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Question: Consider the charges q, q and –q placed at the vertices of an equilateral triangle of each side. l. ...

Consider the charges q, q and –q placed at the vertices of an equilateral triangle of each side. l. The force on the system of charges is.

A

q24πε0l\frac{q^{2}}{4\pi\varepsilon_{0}l}

B

q34πε0l\frac{q^{3}}{4\pi\varepsilon_{0}l}

C

q24πε0l2\frac{q^{2}}{4\pi\varepsilon_{0}l^{2}}

D

Zero

Answer

Zero

Explanation

Solution

: From diagram, force on q1(=q)q_{1}( = q) at A,

F→1=F→12+F→13=Fr^l{\overrightarrow{F}}_{1} = {\overrightarrow{F}}_{12} + {\overrightarrow{F}}_{13} = F{\widehat{r}}_{l}

HereF=q24πε0l2F = \frac{q^{2}}{4\pi\varepsilon_{0}l^{2}} and r^\widehat{r}, is the unit vector along BC

Force on q2(=q)atB,q_{2}( = q)atB,

F→2=F→21+F→23=Fr^2{\overrightarrow{F}}_{2} = {\overrightarrow{F}}_{21} + {\overrightarrow{F}}_{23} = F{\widehat{r}}_{2}

(here r^2,{\widehat{r}}_{2,}is the unit vector along AC) Force on q3(=−q)atCq_{3}( = - q)atC F→3=F→31+F→32=(F12+F22+2F1F2cos⁡60o)n^{\overrightarrow{F}}_{3} = {\overrightarrow{F}}_{31} + {\overrightarrow{F}}_{32} = \left( \sqrt{F_{1}^{2} + F_{2}^{2} + 2F_{1}F_{2}\cos 60^{o}} \right)\widehat{n}

=3Fn^= \sqrt{3}F\widehat{n}

Here n^=\widehat{n} =unit vector along the direction bisecting <BCA

Or F→1+F→2+F→3=0{\overrightarrow{F}}_{1} + {\overrightarrow{F}}_{2} + {\overrightarrow{F}}_{3} = 0