छह बिंदु आवेश R त्रिज्या के एक वृत्त की परिधि पर एक-दूसरे से \( 60^{\circ} \) के कोणीय अंतराल पर रखे गए हैं जैसा कि चित्र में दर्शाया गया है। वृत्त के केंद्र पर कुल विद्युत क्षेत्र __________ है।
(\( \epsilon \) निर्वात की विद्युतशीलता है)

A \( -\frac{5Q}{8\pi\epsilon_{o}R^{2}}(\hat{i}+\sqrt{3j}) \)
B \( -\frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{i}-\hat{j}) \)
C \( -(\frac{5Q}{8\pi\epsilon_{0}R^{2}})(\hat{i}-3\hat{j}) \)
D \( \frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{i}-\hat{j}) \)

Verified Solution

Answer & Solution

Correct Answer

(B) \( -\frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{i}-\hat{j}) \)

Step-by-step Solution

Detailed explanation

माना \( \frac{KQ}{r^{2}}=E_{0} \) \( \vec{E}_{net}=2E_{0}cos~30^{\circ}(-\hat{i})+2E_{0}sin~30^{\circ}(\hat{j}) \) \( =\frac{2kQ}{r^{2}}[\frac{\sqrt{3}}{2}(-\hat{i})+\frac{1}{2}\hat{j}] \) \( =\frac{-1Q}{4\pi\epsilon_{0}{r^{2}}}(\sqrt{3}\hat{i}-\hat{j}) \)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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