The calculated electric field is 3.0 N/C
Explanation:
The electric field is described only as the charge that it produces and is unique at every point in space. In particular, the electric field E is defined as the ratio of Coulomb forces to test charge. It can express as follows,
[tex]E=\frac{Force}{q}[/tex]
Where, F is the electrostatic (or Coulomb) force exerted on the positive test charge q. It is clear that E acts in the same direction of force. It is also assumed that q is so small and will not change the distribution of the charge generated by the electric field. It can express by a unit Newton per Coulomb (N / C).
Here, the given data, q = 2.0 C and F = 6.0 N. So, the electric field would be,
[tex]E = \frac{6.0}{2.0} = 3.0 \mathrm{N} / \mathrm{C}[/tex]
