Answer:
The magnitude of force is [tex]4.26\times 10^{- 6} N[/tex]
Solution:
As per the question:
The strength of Electric field due west at a certain point, [tex]\vec{E_{w}} = 710,000 N/C[/tex]
Charge, Q = - 6 C
Now, the force acting on the charge Q in the electric field is given by:
[tex]\vec{F} = Q\vec{E_{w}}[/tex]
[tex]\vec{F} = -6\times 710,000 = - 4.26\times 10^{- 6} N[/tex]
Here, the negative sign indicates that the force acting is opposite in direction.
