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63 rd question in the attachment.......................................
3 months ago

Arun
23760 Points

Dear student

$\dpi{80} E=-\dfrac{dV}{dx}=\dfrac{1}{2}\dfrac{\lambda r}{2\epsilon_0 (x^2+r^2)^{3/2}}.2x=\dfrac{\lambda rx}{2\epsilon_0(x^2+r^2)^{3/2}}$Hence$\dpi{80} V=\dfrac{1}{4\pi\epsilon_0}\dfrac{\int dq}{\sqrt{x^2+r^2}}=\dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda( 2\pi r)}{\sqrt{x^2+r^2}}=\dfrac{\lambda r}{2\epsilon_0\sqrt{x^2+r^2}}$Hence total electric potential at that point=$\dpi{80} d=\sqrt{x^2+r^2}$Distance of each infinitesimally small charge element on the ring from the given point is

Hope it helps
3 months ago
Vikas TU
10469 Points
3 months ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

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