Answer:
[tex]2.35\cdot 10^{-15} N[/tex]
Explanation:
When a charge is in an electric field, it experiences a force given by
[tex]F=qE[/tex]
where
q is the charge
E is the strength of the electric field
The field between two parallel plates is uniform, so it can be rewritten as
[tex]E=\frac{V}{d}[/tex]
where
V is the potential difference between the plates
d is the separation between the plates
So the first equation becomes
[tex]F=\frac{qV}{d}[/tex]
In this problem we have:
[tex]q=2e=2(1.6\cdot 10^{-19}C)=3.2\cdot 10^{-19}C[/tex] is the magnitude of the charge
V = 250 V is the potential difference between the plates
d = 3.4 cm = 0.034 m is the separation between the plates
So the magnitude of the force is
[tex]F=\frac{(3.2\cdot 10^{-19})(250)}{0.034}=2.35\cdot 10^{-15} N[/tex]
