contestada

If the total positive charge is Q = 1.62×10−6 C , what is the magnitude of the electric field caused by this charge at point P, a distance d = 1.53 m from the charge?

Respuesta :

Answer:

[tex]E = 6228.374\,\frac{N}{C}[/tex]

Explanation:

A definition of electric field for a particle comes from dividing electrostatic force by a hypothetical charge [tex]q_{o}[/tex]. In other words:

[tex]E = \frac{F}{q_{o}}[/tex]

[tex]E = \frac{k\cdot q}{d^{2}}[/tex]

[tex]E = \frac{(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (1.62 \times 10^{-6}\,C)}{(1.53\,m)^{2}}[/tex]

[tex]E = 6228.374\,\frac{N}{C}[/tex]