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`        A THIN HALF RING OF RADIUS 20cm IS UNIFORMLY CHARGE WITH A TOTAL CHARGE Q=0.7nC.THE MAGNITUDE OF ELECTRIC FIELD AT THE CENTRE OF THE HALF RING IS`
2 years ago

Vikas TU
9484 Points
```							We would need to calculate the electric field in both direction that is x and y respeectively.Therefore for x direcn.,dE = k’*k*dl*cos(thetha)/r^2where k’ be the charge densite over the semi arc.Now integerate both sides from theta 0 to pi.O integerating we get,Ex = (k*k’/r)(sinpi – sin0) = 0Similarly in y direcn. we get,dEy= k’*k*dl*cos(thetha)/r^2Integerating both sides,Ey = 2*K*K’/r Net electric field => E = Ey => 2KK’/rwhere K = 1/4pieand K’ = 0.7 x 10^-9/pi*0.2.
```
2 years ago
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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