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When placed at a certain point, a 4.20 C charge feels an electric force of 2.55 N. What is the magnitude of the electric field at that point?



Answer :

To determine the magnitude of the electric field at a specific point where a charge is experiencing an electric force, we use the basic relationship between electric force ([tex]\(F\)[/tex]), charge ([tex]\(q\)[/tex]), and electric field ([tex]\(E\)[/tex]):

[tex]\[ E = \frac{F}{q} \][/tex]

Here, [tex]\(E\)[/tex] represents the electric field, [tex]\(F\)[/tex] represents the electric force acting on the charge, and [tex]\(q\)[/tex] represents the magnitude of the charge itself.

1. Identify the given values:
- The charge ([tex]\(q\)[/tex]) is [tex]\(4.20\)[/tex] Coulombs (C).
- The electric force ([tex]\(F\)[/tex]) is [tex]\(2.55\)[/tex] Newtons (N).

2. Substitute these values into the formula:
[tex]\[ E = \frac{F}{q} \][/tex]
[tex]\[ E = \frac{2.55\ \text{N}}{4.20\ \text{C}} \][/tex]

3. Perform the division to find the electric field:
When you divide [tex]\(2.55\)[/tex] by [tex]\(4.20\)[/tex], you get:
[tex]\[ E \approx 0.6071\ \text{N/C} \][/tex]

Therefore, the magnitude of the electric field at that point is approximately [tex]\(0.6071\)[/tex] Newtons per Coulomb (N/C).