WBJEE · Physics · Electrostatics
Two positive charges \(Q\) and \(4 Q\) are placed at points \(A\) and \(B\) respectively, where \(B\) is at a distance \(d\) units to the right of \(A\). The total electric potential due to these charges is minimum at \(P\) on the line through \(A\) and \(B\). What is (are) the distance (s) of \(P\) from \(A ?\)
- A \(\frac{d}{3}\) units to the right of \(A\)
- B \(\frac{d}{3}\) units to the left of \(A\)
- C \(\frac{d}{5}\) units to the right of \(A\)
- D \(d\) units to the left of \(A\)
Answer & Solution
Correct Answer
(A) \(\frac{d}{3}\) units to the right of \(A\)
Step-by-step Solution
Detailed explanation
\(\because V_{P}\) is minimum. \(\therefore\) \(E_{p}=0 \Rightarrow E_{A}=E_{B}\) \(\Rightarrow \quad \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q}{(d-r)^{2}}\)…
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