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A positively charged particle of charge ...

A positively charged particle of charge q and mass m is suspended from a point by a string of length l. In the space a uniform horizontal electric field E exists . The particle is drawn aside so that it is projected horizontally with velocity v such that the particle starts to move along a circle with same constant speed v. Find the speed v.

A

`(2qE)/m(sqrtl/g)`

B

`(qE)/m(sqrtl/sqrtg)`

C

`(qE)/m(sqrt(2l)/g)`

D

`(qE)/(2m)(sqrtl/g)`

Text Solution

Verified by Experts

The correct Answer is:
B
b. `theta` is equilibrium position. Net force in equilibrium position
`F=sqrt((mg)^(2)+(qE)^(2))`
`tantheta=(qE)/(mg)`
`Tcostheta=F`
`Tsintheta=(mv^(2))/(r)`
where `r=lsin theta`
Solve to get `v=(qE)/(m) sqrt((l)/(g))`
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