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Grade: 12
        
63 rd question in the attachment.......................................
3 months ago

Answers : (2)

Arun
23760 Points
							
Dear student
 
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}}HenceV=\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=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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