In this case, the force is $6.0 \times 10^{-6} \mathrm{~N}$ and the charge is $-2.0 \times 10^{-9} \mathrm{C}$. So, the magnitude of the electric field is given by:
\[E = \frac{F}{q} = \frac{6.0 \times 10^{-6} \mathrm{~N}}{-2.0 \times 10^{-9} \mathrm{C}} = 3.0
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