TS EAMCET · Physics · Electrostatics
Two point charges \(-10 \mu \mathrm{C}\) and \(+5 \mu \mathrm{C}\) are situated on \(X\)-axis at \(x=0\) and \(x=\sqrt{2} \mathrm{~m}\). The point along the \(X\)-axis where the electric field becomes zero is
- A \(x=(\sqrt{2}-1) \mathrm{m}\)
- B \(x=2(\sqrt{2}-1) \mathrm{m}\)
- C \(x=2(\sqrt{2}+1) \mathrm{m}\)
- D \(x=(\sqrt{2}+1) \mathrm{m}\)
Answer & Solution
Correct Answer
(C) \(x=2(\sqrt{2}+1) \mathrm{m}\)
Step-by-step Solution
Detailed explanation
The point on x -axis where electric field is zero, \(\begin{aligned} & x=\left(\frac{\sqrt{q_2}}{\sqrt{q_1}-\sqrt{q_2}}\right) r=\left(\frac{\sqrt{5}}{\sqrt{10}-\sqrt{5}}\right) \sqrt{2} \\ & =\frac{\sqrt{2}}{\sqrt{2}-1} \mathrm{~m} \end{aligned}\) \(\therefore\) Distance of…
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