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A charged ball of mass 9 kg is suspended...

A charged ball of mass 9 kg is suspended from a string in a uniform elelctric field `vecE = (3hati + 5hatj) xx 10^5 N//C`. The ball is in equilibrium with `theta=37^@`. If direction of electric field is reversed, find the new equilibrium position of the ball. Given your answer in terms of angle made by string with vertical. Take `g = 10ms^(-2)`.

A

`tan^(-1)(3/4)`

B

`cot^(-1) (3/14)`

C

`cot^(-1) (3/4)`

D

`tan^(-1)(3/14)`

Text Solution

Verified by Experts

The correct Answer is:
D

`Tsin theta=3thetaxx10^(5)` (i)
`Tcostheta=mg-5qxx10^(5)` (ii)
Solve to get `q=100muC,T=50N`.
After the rever4sal of direction of electric field

`T'sin a=3qxx10^(5)` or `T'cos a=mg+5qxx10^(5))`
`tana=(3qxx10^(5))`
`=(3xx10^(-4)xx10^(5))/(9xx10+5xx10^(-4)xx10^(5))=(3)/(14)`
or `a=tan^(-1)((3)/(14))`
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