Uniform Electric Field Upward Electric Force Equals Weight. HELP

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An object with a charge of -3.6 µC and a mass of 0.012 kg experiences an upward electric force equal to its weight, leading to calculations for the electric field. The weight is calculated as 0.12 N, and using the formula W = kQ/r², the radius r is determined to be approximately -431.52. The electric field is then calculated to be -2.78 x 10^-4 N/C in the y direction. For part b, when the charge is doubled, the acceleration is found to be 9.81 m/s², indicating that the electric force on the charge is equal to the gravitational force. The relationship between electric field strength and charge is confirmed by the equation E*q = mg.
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1. An object with a charge of -3.6uC and a mass of .012kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. (a) Find the direction and magnitude of the electric field. (b) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.
2. E=kQ/r(sqrd)
k=8.99 x 10^9 N*m^2/C^2
magnitude of electron's charge (e): 1.60 x 10 ^ -19 C

3. Attempt
W=mg
W= (.012kg)(10)
W=.12N

Upward force equal to magnitude in weight so...
W= W=kQ/r(sqrd)
.12N + (8.99 x 10^9) [(-3.6uC)(1.6 x 10^-19])/(r^2)
Solve for r
r=-431.52

Plug r into E equation to find electric charge.
E=kQ/r(sqrd)
E=(8.99 x 10^9) [(-3.6uC)(1.6 x 10^-9)]/(-431.52^2)
E= -51.7824/-431.52^2
E= -2.78 x 10 ^ -4

I have no idea about part b. Is it always equal to gravity? Is that how they found the answer.

Correct answer from book:
a) (-3.3 x 10^4 N/C) y direction
b) (9.81 m/s^2) y direction
 
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In the electric field force on a charge is equal to E*q
Hence mg = E*q. Now solve for E.
 
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