If a particle with a charge of [tex]+3.3 \times 10^{-18} C[/tex] is attracted to another particle by a force of [tex]2.5 \times 10^{-8} N[/tex], what is the magnitude of the electric field at this location?

A. [tex]8.3 \times 10^{-26} \, \text{N/C}[/tex]
B. [tex]1.8 \times 10^{10} \, \text{N/C}[/tex]
C. [tex]1.3 \times 10^{-10} \, \text{N/C}[/tex]
D. [tex]7.6 \times 10^9 \, \text{N/C}[/tex]



Answer :

To determine the magnitude of the electric field at the location where a particle with a charge [tex]\(+3.3 \times 10^{-18} \text{ C}\)[/tex] experiences a force of [tex]\(2.5 \times 10^{-8} \text{ N}\)[/tex], we can use the relationship between the electric field [tex]\(E\)[/tex], the force [tex]\(F\)[/tex], and the charge [tex]\(q\)[/tex]. This relationship is given by the formula:

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

Where:
- [tex]\(E\)[/tex] is the magnitude of the electric field,
- [tex]\(F\)[/tex] is the force experienced by the charge,
- [tex]\(q\)[/tex] is the charge.

Given the values:
- [tex]\(F = 2.5 \times 10^{-8} \text{ N}\)[/tex]
- [tex]\(q = +3.3 \times 10^{-18} \text{ C}\)[/tex]

We can plug these values into the formula to find [tex]\(E\)[/tex]:

[tex]\[ E = \frac{2.5 \times 10^{-8}}{3.3 \times 10^{-18}} \][/tex]

By performing the division, we get:

[tex]\[ E \approx 7575757575.757575 \text{ N/C} \][/tex]

Thus, the magnitude of the electric field at this location is approximately [tex]\(7.6 \times 10^9 \text{ N/C}\)[/tex].

Therefore, the correct answer is:

[tex]\[ \boxed{7.6 \times 10^9 \text{ N/C}} \][/tex]