Answer:
E = -477 N/C
Explanation:
Given that,
Charge, [tex]q=-4.77\times 10^{-9}\ C[/tex]
We need to find the electric field at a point 0.3 m to the right of the charge.
We know that the electric field at a distance r is given by :
[tex]E=\dfrac{kq}{r^2}\\\\E=\dfrac{9\times 10^9\times 4.77\times 10^{-9}}{(0.3)^2}\\\\E=-477\ N/C[/tex]
So, the electric field is -477 N/C.
