Two small plastic balls hang from threads (1m long) of negligiblemass. Each ball

Question

Two small plastic balls hang from threads (1m long) of negligiblemass. Each ball has a mass of 0.09g and equal charges. The ballsare repelled by each other .they are 18cm apart and the threadsattached to the balls make an angle of 25 degree with the vertical. Determine:
i) the magnitude of the electric force acting on eachball. ii) the tension in each of the threads. iii) the magnitude of the charge q on the balls.
i) the magnitude of the electric force acting on eachball. ii) the tension in each of the threads. iii) the magnitude of the charge q on the balls.

Explanation / Answer

   mass of the each ball m = 0.90g=0.90*10-3kg    Acceleration due to gravity g = 9.8m/s2    k =8.9*10^9 N.m2 /C2
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The electric force between two balls is given by            F = (8.9 *10^9N.m2/C2) (q2)/(0.18m)2   -------------(1) also            F = T sin 25 ----------(2)
         T cos25 =mg              T= ( mg) /cos25               =((0.90*10-3kg)( 9.8 m/s2))/cos25    Simplifying we get T i.e tension in thestring.
From (2)      (8.9*10^9N.m2/C2) (q2)/(0.18m)2  = T sin 25 solve for q i.e charge on the ball.
substitute the values of q in (1) we get electric forceF. The electric force between two balls is given by            F = (8.9 *10^9N.m2/C2) (q2)/(0.18m)2   -------------(1)            F = T sin 25 ----------(2)
         T cos25 =mg              T= ( mg) /cos25               =((0.90*10-3kg)( 9.8 m/s2))/cos25    Simplifying we get T i.e tension in thestring.
From (2)      (8.9*10^9N.m2/C2) (q2)/(0.18m)2  = T sin 25 solve for q i.e charge on the ball.
substitute the values of q in (1) we get electric forceF.

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