Answer:
Electric field, [tex]E=5.63\times 10^{16}\ N/C[/tex]
Explanation:
Given that,
Charge, [tex]q=1.83\times 10^5\ C[/tex]
We need to find the magnitude of electric field 17.1 cm (0.171 m) above an isolated charge. Electric field at a point is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]E=\dfrac{9\times 10^9\times 1.83\times 10^5\ C}{(0.171\ m)^2}[/tex]
[tex]E=5.63\times 10^{16}\ N/C[/tex]
So, the electric field is [tex]5.63\times 10^{16}\ N/C[/tex]. Hence, this is the required solution.
