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A pendulum bob of mass m = 80 mg, carryi...

A pendulum bob of mass m = 80 mg, carrying a charge of `q=2xx10^(-8)C`, is at rest in a horizontal, uniform electric field of E = 20,000 V/m. The tension T in the thread of the pendulum and the angle `alpha` it makes with vertical, is (take `g=9.8m//s^(2)`)

A

`3.92xx10^(-8),72^(@)`

B

`7.84xx10^(-6),90^(@)`

C

`8.8xx10^(-4),27^(@)`

D

`1.96xx10^(-4),18^(@)`

Text Solution

Verified by Experts

The correct Answer is:
c
the different forces acting on the pendulum bob are :
(i) Weight of the bob , mg
(ii) Tension in the thread of the bob , T and
(iii) Force on the charged charged bob due to the electric field, F . Here , Mass of the bob ,
`m=80mg=80xx10^(-6)kg`
Charge on the bob,
`q=2xx10^(-8)` coulomb Electric field,
`E=20,000V//m`
`thereforeF=qE=2xx10^(-8)xx20,000N`
Let the thread makes an angle `theta` with the vertical at its equilibrium position.
Horizontal and vertical force balances of the bob give
`Tsintheta=F` and `Tcostheta=mg` ..(ii)
Dividing (i) by (ii) , we get
`tantheta=(F)/(mg)=(2xx10^(-8)xx20,000)/(80xx10^(-6)xx9.8)["taking"g =9.8m//s^(2)]`
`=0.51`
`thereforetheta=tan^(-1)(0.51)=27^(@)`
From (ii) , we have
`T=(mg)/(costheta)=(80xx10^(-6)xx9.8)/(cos27^(@))=8.8xx10^(-4)N`
Therefore , the tension in the thread is `8.8xx10^(-4)` N and it makes an angle `27^(@)` with the vertical
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