Answer:
[tex]q=-1.96\cdot 10^{-7}C[/tex]
Explanation:
In order for the ball to be suspended, the electric force on it must be equal to the weight of the ball, so we can write:
[tex]qE=mg[/tex]
where
qE is the electric force
mg is the weight
q is the charge on the ball
E = 6000 N/C is the electric field
[tex]m=0.120 g = 0.120 \cdot 10^{-3} kg[/tex] is the mass of the ball
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for q, we find the magnitude of the charge:
[tex]q=\frac{mg}{E}=\frac{(0.120\cdot 10^{-3})(9.8)}{6000}=1.96\cdot 10^{-7} C[/tex]
Now we have to determine the direction. We know that:
- The weight points downward, so the electric force must acts upwnward
- The electric field points downward: so, F and E have opposite directions
- This means that the charge is negative
So, the answer is
[tex]q=-1.96\cdot 10^{-7}C[/tex]