Four point charges, each of + q, are rigidly | vedclass

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Question

$F_1=$ Net electrostatic force on any one charge due to rest of three charges

$=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}\left(\sqrt{2}+\frac{1

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$F_1=$ Net electrostatic force on any one charge due to rest of three charges

$=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}\left(\sqrt{2}+\frac{1}{2}ight), F_2=$ surface this equation we get $a =2$.

If we see the equilibrium of line $B C$, then .

$2 F_1 \cos 45^{\circ}=F_2 	ext { or } \sqrt{2} F_1=F_2 	ext { or } \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}(2 +\frac{1}{\sqrt{2}}=\gamma a$

$	herefore a^3=\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}} \frac{q^2}{\gamma} 	ext { or } aight.$

$=\left\{\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}ight\}^{1 / 3}\left[\frac{q^2}{\gamma}ight)ight]^{2 / k}=k\left[\frac{q^2}{\gamma}ight]^{1 / 3}$

Where $k=\left\{\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{2}ight)ight\}^{1 / 3}$ therefore $N =3$

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