Answer:
[tex]4,000N/C[/tex]
Explanation:
the information we have is:
distance: [tex]r=3m[/tex]
charge: [tex]q= 4\mu C[/tex]
since the prefix [tex]\mu[/tex] is equal to [tex]1x10^{-6}[/tex]
the charge is: [tex]q=4x10^{-6}C[/tex]
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To calculate the strength of the electric field we use the following formula:
[tex]E=\frac{kq}{r^2}[/tex]
where [tex]q[/tex] is the charge, [tex]r[/tex] is the distance and [tex]k[/tex] is Coulomb's constant
[tex]k=9x10^9Nm^2/C^2[/tex]
substituting all the known values into the formula:
[tex]E=\frac{(9x10^9Nm^2/C^2)(4x10^{-6}C)}{(3m)^2} \\E=\frac{36,000Nm^2/C}{9m^2} \\E=4,00N/C[/tex]
The strength of the electric field on the point charge at the distance of 3m is 4,000 N/C.
