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Figure shows a rod AB, which is bent in ...

Figure shows a rod AB, which is bent in a `120^(@)` circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field `vecE` at the centre of curvature O ?

A

`(3sqrt(3)Q)/(8pi^(2)epsilon_(0)R^(2))(hati)`

B

`(3sqrt(3)Q)/(8pi^(2)epsilon_(0)R^(2))(-hati)`

C

`(3sqrt(3))/(8piepsilon_(0)R^(2))(hati)`

D

`(3sqrt(3)Q)/(16pi^(2)epsilon_(0)R^(2))(hati)`

Text Solution

Verified by Experts

The correct Answer is:
A

`=underset(-pi//3)overset(pi//2)int(Kxx(+Q))/((2pi)/(3)R)xx(Rd theta)/(R^(2))xxcostheta`
`=(3)/(2pi)(KQ)/(R^(2))[sintheta]_(-pi//3)^(pi//3)=(3)/(2pi)(KQ)/(R^(2))xx(2sqrt(3))/(2)`
`therefore vecE=(3sqrt(3))/(8pi^(2)epsilon_(0)R^(2))(hati)`.
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