JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
An electric charge \(10^{-6} \mu \mathrm{C}\) is placed at origin \((0,0)\) \(\mathrm{m}\) of \(\mathrm{X}-\mathrm{Y}\) co-ordinate system. Two points \(\mathrm{P}\) and \(\mathrm{Q}\) are situated at \((\sqrt{3}, \sqrt{3}) \mathrm{m}\) and \((\sqrt{6}, 0) \mathrm{m}\) respectively. The potential difference between the points \(P\) and \(Q\) will be :
- A \(\sqrt{3} \mathrm{~V}\)
- B \(\sqrt{6} \mathrm{~V}\)
- C \(0 \mathrm{~V}\)
- D \(3 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(0 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Potential difference \(=\frac{K Q}{r_1}-\frac{K Q}{r_2}\) \( \mathrm{r}_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2}\) \( \mathrm{r}_2=\sqrt{(\sqrt{6})^2+0}\) As \(r_1=r_2=\sqrt{6} \mathrm{~m}\) So potential difference \(=0\)
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