78. What is the magnitude of the electric field at a field point 2.0 m from a point charge q = 4.0 nC? (a) 18.0 N/C (b) 9.0 N/C (c) 36.0 N/C (d) 6.0 N/C
Answer (1)
To determine the magnitude of the electric field at a point 2.0 meters away from a point charge, we can use Coulomb's Law for the electric field due to a point charge.
The electric field E due to a point charge q is given by the equation:
E = r 2 k ∣ q ∣
where:
k is Coulomb's constant, approximately 8.99 × 1 0 9 N m 2 / C 2 .
q is the charge, in this case 4.0 nC , which is 4.0 × 1 0 − 9 C .
r is the distance from the charge, which is 2.0 m .
Substitute the values into the equation:
E = ( 2.0 m ) 2 8.99 × 1 0 9 N m 2 / C 2 × 4.0 × 1 0 − 9 C
E = 4 8.99 × 4.0 × 1 0 0 N/C
E = 8.99 × 1.0 N/C = 8.99 N/C
Therefore, after rounding, the magnitude of the electric field is approximately 9.0 N/C .
Hence, the correct answer is option (b) 9.0 N/C .
