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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, fing the new equilibrium position of the ball. Given your answer in terms of angle made by string with vertical. Take g = 10ms^(-2). |
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Answer» `TAN^(-1)(3/4)`  `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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