AP EAMCET · PHYSICS · Electrostatics
Let \(\mathrm{V}_1\) be the potential at the center of the square of side \(1 \mathrm{~m}\) when the charges at the 4 corners are \(2 \mathrm{C}\) each. If the same charges are placed at the corners of a square of side \(2 \mathrm{~m}\), then the potential at the centre of this square is \(\mathrm{V}_2\). The value of \(\frac{V_2}{V_1}\) is
- A \(\frac{1}{2}\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\frac{1}{2 \sqrt{2}}\)
- D \(\frac{1}{4 \sqrt{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\left(\mathrm{V}_{\text {centre }}\right)_1=\frac{\mathrm{K} \times 2}{\frac{1}{\sqrt{2}}} \times 4=\mathrm{V}_1\) \[ \left(\mathrm{V}_{\text {centre }}\right)_2=\frac{\mathrm{K} \times 2}{\frac{2}{\sqrt{2}}} \times 4=\mathrm{V}_2 \] So,…
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