A 25 nC charge experiences a 0.047 N electric force.What is the magnitude of electric field at the position of this charge?

Question

asked Aug 7, 2024 169k views

1 Answer

Bublik · Answer 1 · 2024-08-12T09:35:22+0000

The magnitude of the electric field at the position of the charge is 1.88 x 10^6 N/C.

The magnitude of the electric field at the position of the charge can be found using the equation:

E = F / q,

where E is the electric field, F is the electric force, and q is the charge. Plugging in the given values:

E = 0.047 N / 25 x 10^-9 C = 1.88 x 10^6 N/C.

Therefore, the magnitude of the electric field at the position of the charge is 1.88 x 10^6 N/C.