Question

calculate the electric field intensity at a point 3m distance from a 6C in vacuum​

Answer 1

$electric \: field \: intensity \: (e) = \frac{q}{4\pi \times epsilon \times {r}^{2} }$

where epsilon is the permittivity of air (vacuum) and q=6 C

Answer 2

Answer:

The electric field intensity at a point $3m$ distance from a $6c$ in vacuum​ is$28.2N/C$

Explanation:

The force that one charged body is exerting on another is measured by the intensity of the electric field. According to tradition, imaginary lines of force or electric field lines start on positive charges and end on negative charges.

This electric field strength formula's illustration of the inverse square relationship between electric field strength and distance is one of its key features. The square of the distance from the source has an inverse relationship with the strength of the electric field produced by source charge Q. An inverse square law applies in this situation.

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