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        63 rd question in the attachment.......................................
one month ago

## Answers : (2)

Arun
23520 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

one month ago
Vikas TU
10070 Points
							Dear student Please follow this link https://www.askiitians.com/forums/Electrostatics/consider-a-circular-ring-of-radius-r-uniformly-c_218006.htmGood Luck

one month ago
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### Course Features

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### Course Features

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