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A non-conducting hollow cone has charge ...

A non-conducting hollow cone has charge density `sigma` A path `ABP` is cut and removed from the cone. The potential due to the remaining portion of the cone at point `'P'` is

A

`5/6 (sigmaR)/(epsilon_(0))`

B

`5/24 (sigmaR)/(epsilon_(0))`

C

`5/3 (sigmaR)/(epsilon_(0))`

D

`5/12 (sigma R)/(epsilon_(0))`

Text Solution

Verified by Experts

The correct Answer is:
D
Potential due to circular non conducting disc, `V_(c)=(sigmar)/(2epsilon_(0))` (where `r` is radius of disc)
So `v_(P)=((theta)/(2pi))((sigma.l)/(2epsilon_(0)))`
`=((sigmal)/(2epsilon_(0)))(5/3 (piR)/((l)))(1/(2pi)=(5sigmaR)/(12 epsilon_(0))`
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