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        'wo identical thin rings, each ofradiuses X are coaxially placed at a distance,ll apart. lt e1 and p2 are respectively the charges uniformly spread on the two rings. The work done in moving a charge g fronr the centre ofone ring to that ofthe second ring is
6 months ago

Arun
22956 Points
							E and V at a point P that lies on the axis of ring -$\dpi{100} E_{x}=\frac{kQx}{\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}$  ,   $\dpi{100} V=\frac{kQ}{\left ( x^{2}+R^{2} \right )^{\frac{1}{2}}}$-  Charged Circular ring -A charged circular ring of radius R and charge Q.- wherein    Potential at the centre of first ring $V_{A}=\frac{Q_{1}}{4\pi \varepsilon _{0}R}+\frac{Q_{2}}{4\pi \varepsilon _{0}\sqrt{R^{2}+R^{2}}}$Potential at the centre of second ring $V_{B}=\frac{Q_{2}}{4\pi \varepsilon _{0}R}+\frac{Q_{1}}{4\pi \varepsilon _{0}\sqrt{R^{2}+R^{2}}}$Potential difference between the two centres $V_{A}-V_{B}=\frac{\left ( \sqrt{2}-1 \right )\left ( Q_{1}-Q_{2} \right )}{4\pi \varepsilon _{0}R\sqrt{2}}$$\therefore Work\; done$     $W=\frac{q\left ( \sqrt{2}-1 \right )\left ( Q_{1}-Q_{2} \right )}{4\pi \varepsilon _{0}R\sqrt{2}}$

6 months ago
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• 728 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions