Question

80PTS

what is the magnitude and direction of the electric field 4.52 to the left of a 8.15 c negative charge?
PLZ SHOW WORK

Answer 1

Answer: what the person said on the top it correct

Step-by-step explanation:

Answer 2

a) The magnitude of the electric field generated by a single point charge is given by

`E=k (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 (q)/(r^2)=(8.99 \cdot 10^9 Nm^2C^(-2)) (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.