Respuesta :
Answer : Charge, [tex]q=3.0\ \muC[/tex]
Explanation :
It is given that :
Electric field, [tex]E=4286\ N/C[/tex]
Distance, [tex]r=2.5\ m[/tex]
We know that the electric field at any point is given by :
[tex]E=k\dfrac{q}{r^2}[/tex]
k is the electrostatic constant.
q is the electric charge
r is the distance.
[tex]q=\dfrac{E\times r^2}{k}[/tex]
[tex]q=\dfrac{4286\ N/C\times (2.5\ m)^2}{9\times 10^9\ Nm^2/C^2}[/tex]
[tex]q=0.0000029\ C[/tex]
[tex]q=2.9\ \mu C[/tex]
or
[tex]q=3.0\ \mu C[/tex]
The correct option is (b).
Hence, this is the required solution.