Two point charges q 1 = + 2C and q 2 = – 1C are separated by a distance d. The position…
Physics · JEE Main · NTA Exams — Electrostatics
Two point charges q1 = + 2C and q2 = – 1C are separated by a distance d. The position on the line joining the two charges where a third charge = + 1C will be in equilibrium is at a distance
- \(\mathrm{d} / \sqrt{2}\)from q1 between q1 & q2
- \(\mathrm{d} / \sqrt{2}\)from q1 away from q2
- \(\mathrm{d} / \sqrt{2}-1\) from q2 between q1 & q2
- \(\mathrm{d} / \sqrt{2}-1\) from q2 away from q1
Answer
(D) d / 2 -1 from q 2 away from q 1
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