The magnitude of the electric field is [tex]1.79\cdot 10^{-5} N/C[/tex], the direction is to the right
Explanation:
The magnitude of the electric field produced by a single point charge is given by
[tex]F=k\frac{q}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q[/tex] is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
[tex]q=6.7\cdot 10^{-16}C[/tex] is the charge
r = 58 cm = 0.58 m is the distance of point P from the charge
Substituting, we find the magnitude of the field at point P:
[tex]E=(8.99\cdot 10^9) \frac{6.7\cdot 10^{-16}}{0.58^2}=1.79\cdot 10^{-5} N/C[/tex]
The direction of the field for a positive charge is away from the charge: therefore, since point P is to the right of the charge, the direction of the field is to the right.
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