The calculated electric field is 3.0 N/C
Step-by-step 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,
![E=(Force)/(q)](https://img.qammunity.org/2020/formulas/physics/middle-school/wy7667d9uh2a36sal7fvprcvpu3rhama6y.png)
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,
![E = (6.0)/(2.0) = 3.0 \mathrm{N} / \mathrm{C}](https://img.qammunity.org/2020/formulas/physics/middle-school/3xhl6uv2rwuwkih53da1us06xp71bnluen.png)