Question

80PTS

Answer 1

A) The magnitude of the electric field generated by a single point charge is given by
$E=k \frac{q}{r^2}$
where k is the Coulomb's constant, q is the charge and r is the distance at which we calculate the field.
Since we want to know the intensity of the field at distance of r=4.52 m from the charge of q=8.15 C, the intensity of the electric field is
$E=k \frac{q}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{8.15 C}{4.52 m}=3.6 \cdot 10^9 N/C$

b) Direction of the field
The charge that produces the field is negative, and we know that the electric field generated by a negative charge points toward the charge. Therefore, the direction of the electric field is towards the charge.