m द्रव्यमान के एक बिन्दु आवेश q को ell लम्बाई | vedclass

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Question

$U _{ i }=0$

$U _{ f }=\frac{ kq P }{\left(2 \ell \sin \frac{\alpha}{2}ight)^2}+ mgh$

Now, from $	riangle OAB$

$\alpha+90-	heta+90-	heta

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$U _{ i }=0$

$U _{ f }=\frac{ kq P }{\left(2 \ell \sin \frac{\alpha}{2}ight)^2}+ mgh$

Now, from $	riangle OAB$

$\alpha+90-	heta+90-	heta=180$

$\Rightarrow \alpha=2 	heta$

From $	riangle ABC : h =2 \ell \sin \left(\frac{\alpha}{2}ight) \sin 	heta$

$h =2 \ell \sin \left(\frac{\alpha}{2}ight) \sin \left(\frac{\alpha}{2}ight)$

$\Rightarrow  h =2 \ell \sin ^2\left(\frac{\alpha}{2}ight)$

Now charge is in equilibrium at point $B$.

So, using sine rule

$\Rightarrow \frac{ mg }{\sin \left[90+\frac{\alpha}{2}ight]}=\frac{ qE }{\sin [180-2 	heta]}$

$\Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{\sin 2 	heta}$

$\Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{\sin \alpha} \Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}$

$\Rightarrow qE = mg 2 \sin \left(\frac{\alpha}{2}ight)$

$\Rightarrow \frac{ q 2 lq }{\left[2 \ell \sin \frac{\alpha}{2}ight]^3}= mg 2 \sin $$\left(\frac{\alpha}{2}ight) \Rightarrow \frac{ kpq }{\left[2 \ell \sin \frac{\alpha}{2}ight]^2}= mg \sin \left(\frac{\alpha}{2}ight) 	imes\left(2 \ell \sin \frac{\alpha}{2}ight)$

$\Rightarrow \frac{ kpq }{\left[2 \ell \sin \frac{\alpha}{2}ight]^2}= mgh \Rightarrow \operatorname{substituting~this~in~equation~(i)~}$

$\Rightarrow U$

$W =\Delta U = Nmgh = N =2$

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