Answer:
[tex]1.15669\times 10^{-7}\ C[/tex]
Explanation:
[tex]\theta[/tex] = Angle with which the electric field is hung = 23°
m = Mass of ball = 5 g
E = Electric field = [tex]1.8\times 10^5\ N/C[/tex]
T = Tension
q = Charge
We have the equations
[tex]Tcos\theta=mg[/tex]
[tex]Tsin\theta=qE[/tex]
Dividing the equations
[tex]tan\theta=\dfrac{mg}{qE}\\\Rightarrow q=\dfrac{mgtan\theta}{E}\\\Rightarrow q=\dfrac{5\times 10^{-3}\times 9.81\times tan23}{1.8\times 10^5}\\\Rightarrow q=1.15669\times 10^{-7}\ C[/tex]
The magnitude of the charge is [tex]1.15669\times 10^{-7}\ C[/tex]
