Question

Please help! What's the answer to this question, A 5.6 nC electric charge is placed in an Electric Field and experiences a force of 7.4 μN. What is the magnitude of the Electric Field at that location?

Answer 1

Answer:

1321 N/C

Explanation:

From the question,

Electric Field (E) = Electric Force(F)/Electric Charge(q)

E = F/q............ Equation 1

Given: F = 7.4 μN = 7.4×10⁻⁶ N, q = 5.6 nC = 5.6×10⁻⁹ C

Substitute these value into equation 1

E = ( 7.4×10⁻⁶)/(5.6×10⁻⁹)

E = 1.321×10³ N/C

E = 1321 N/C

Hence the magnitude of the electric field at that location is 1321 N/C