An electric field of 790,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -3.00 uC at this spot? (14C = 10 6C) Give your answer in Sl unit rounded to two decimal places.

According to Coulomb's law, the magnitude of electrostatic force of interaction between two static point charges [tex]\rm q_1[/tex] and [tex]\rm q_2[/tex], separated by a distance [tex]\rm r[/tex], is given by

[tex]\rm F = \dfrac{kq_1q_2}{r^2}.[/tex]

where, [tex]k[/tex] = Coulomb's constant.

The direction of this force is along the line joining the two charges, from positive to negative charge.

The electric field at a point is defined as the amount of electrostatic force experienced per unit small positive test charge, placed at that point.

Therefore,

[tex]\rm E = \dfrac Fq\\\\\Rightarrow F = qE.[/tex]

We have,

Electric field, [tex]\rm E = 790,000\ N/C.[/tex]
Charge placed at given spot, [tex]\rm q=-3.00\ \mu C =-3.00\times 10^{-6}\ C.[/tex]

Therefore, the electric force at that point is given by

[tex]\rm F = 3.00\times 10^{-6}\times 790,000=2.37\ N.[/tex]

The negative sign indicates that the force is attractive in nature.

Thus, the magnitude of this force = 2.37 N.