Answer:
[tex]7.6\cdot 10^9 N/C[/tex]
Explanation:
The relationship between force exerted on a charge and strength of the electric field is given by
[tex]F=qE[/tex]
where
F is the strength of the electric force
q is the charge of the particle
E is the magnitude of the electric field
For the particle in the problem, we have
[tex]q=3.3\cdot 10^{-18} C[/tex]
[tex]F=2.5\cdot 10^{-8} N[/tex]
So the magnitude of the electric field at the location of the particle is
[tex]E=\frac{F}{q}=\frac{2.5\cdot 10^{-8}}{3.3\cdot 10^{-18}}=7.6\cdot 10^9 N/C[/tex]
