+91-87964 74404
FlagBACK
Question icon
Grade 12th passAIPMT

A semicircular non conducting ring of radius R is uniformly charged with total charge Q. find the electrostatic force on q placed at the centre of curvature of half ring

Profile image of Vinayak
8 Years agoGrade 12th pass
Answers icon

1 Answer

Profile image of Eshan
8 Years ago
Dear student,

The charge per unit length of the ring=\lambda=\dfrac{Q}{\pi R}
Electric field at the center of curvature of half ring=E=\int_{-\pi/2}^{\pi/2}\dfrac{k(\lambda Rd\theta)}{R^2}cos\theta
=\dfrac{2k\lambda }{R}=\dfrac{\lambda}{2\pi\epsilon_0 R}

Hence force on the charge kept at that point=qE=\dfrac{\lambda q}{2\pi\epsilon_0R}=\dfrac{Qq}{2\pi^2\epsilon_0 R^2}

Other Related Questions on aipmt

question mark
NEET COACHING FOR GRADE 9 STUDENT IN ABU DHABI,UAE. What are the options available
aipmt
1 Answer Available

Last Activity: 4 Years ago

question mark
Mam today in neet ug 2021 due to my nervousness I bubbled one wrong figure of my roll number but I have filled the roll number correctly in the blocks the invigilator adviced me to leave it as it is and to bubble the same correct figure so I did that .am very much worried what will happen
aipmt
1 Answer Available

Last Activity: 4 Years ago

question mark
Magnetic Length of bar magnet of length 70 cm is 63 cm. Want a solution for this.
aipmt
1 Answer Available

Last Activity: 5 Years ago

question mark
in the stomatal apparatus why are subsidiary celss absent in cucurbits?
aipmt
1 Answer Available

Last Activity: 5 Years ago

question mark
Any student studying in Maharashtra since first standard till 10th, takes up NEET coaching in Kota, Rajasthan. Will ge be eligible for admission into MBBS medical colleges in Maharashtra under 85% state  quota seats
 
aipmt
1 Answer Available

Last Activity: 5 Years ago