Answer
The electric force on Object A is $4.5\times 10^{-3}~N$ in the positive y-direction.
The electric force on Object B is $4.5\times 10^{-3}~N$ in the negative y-direction.
Work Step by Step
We can find the magnitude of the electric force on each object:
$F = \frac{k~q_A~q_B}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(10\times 10^{-9}~C)(20\times 10^{-9}~C)}{(0.020~m)^2}$
$F = 4.5\times 10^{-3}~N$
Since one charge is positive and one charge is negative, the two charges attract each other.
The electric force on Object A is $4.5\times 10^{-3}~N$ in the positive y-direction.
The electric force on Object B is $4.5\times 10^{-3}~N$ in the negative y-direction.
