Answer to Question #269501 in Electric Circuits for max

Question #269501
A small 15.0 g plastic ball is suspended by a 30.0 cm string in a uniform electric field of 
3000  N/C, as shown in the picture. If the ball is in equilibrium when the string makes a 45° angle with the vertical as indicated, what is the net charge on the ball?







1
Expert's answer
2021-11-22T10:15:02-0500



Taking E == Intensity of electric field

T== Tension on string

M== Mass of the ball

g== Acceleration due to gravity

q=q= Charge of the plastic ball



Since the ball is equilibrium \sumFy=0F_y=0 And Fx=0\sum\>F_x=0



T sin 45=45= qE

T cos 45=45= mg



Tan 45°=qEmg     q=mgtan45E45°=\frac{qE}{mg}\implies\>q=\frac{mg\>tan45}{E}


Where m=151000=0.015kgm=\frac{15}{1000}=0.015kg

g=9.81g=9.81

E=3000N/CE=3000N/C



q=0.015×9.81tan453000q=\frac{0.015×9.81\>tan45}{3000}


=4.905×105C=4.905×10^{-5}C


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