TS EAMCET · Physics · Electrostatics
A charge q is placed at the centre ' O ' of a circle of radius R and two other charges q and q are placed at the ends of the diameter AB of the circle. The work done to move the charge at point B along the circumference of the circle to a point C as shown in the figure is
- A \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \cdot \frac{\mathrm{q}^2}{\mathrm{R}}(\sqrt{2})\)
- B Zero
- C \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \cdot \frac{\mathrm{q}^2}{\mathrm{R}}\left(\frac{\sqrt{2}-1}{2}\right)\)
- D \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \cdot \frac{\mathrm{q}^2}{\mathrm{R}}\left(\frac{1}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \cdot \frac{\mathrm{q}^2}{\mathrm{R}}\left(\frac{\sqrt{2}-1}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(V_B = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{R} + \frac{q}{2R} \right) = \frac{1}{4 \pi \varepsilon_0} \frac{3q}{2R}\) Distance \(AC = \sqrt{R^2+R^2} = R\sqrt{2}\)…
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