TS EAMCET · Physics · Electrostatics
Two electric charges \(+2 \mu \mathrm{C}\) and \(-4 \mu \mathrm{C}\) are separated by a distance \(3 \mathrm{~m}\) in air. At a point \(\mathrm{P}\) located on the line joining the two charges and in between them, the electric potential is zero. Then the electric field at a point \(\mathrm{P}\left(\right.\) in \(\left.\mathrm{NC}^{-1}\right)\) is
- A 9,000
- B 18,000
- C 12,000
- D 27,000
Answer & Solution
Correct Answer
(B) 18,000
Step-by-step Solution
Detailed explanation
\(\mathrm{Q}_1=+2 \times 10^{-6} \mathrm{C} ; \mathrm{Q}_2=-4 \times 10^{-6} \mathrm{C}\) Distance, \(\mathrm{d}=3 \mathrm{~m}\) Electric potential, \(V=\frac{2 \times 10^{-6}}{4 \pi \epsilon_0(3-x)}+\frac{(-4) \times 10^{-6}}{4 \pi \epsilon_0 x}\)…
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