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4) (7 pts.) A water molecule is centered at the origin of a coordinate system with its dipole moment vector aligned with the x axis. The magnitude of a water molecule dipole is 6.16 × 10−30 C·m. What is the magnitude of the electric field at x = 3.00 × 10−9 m?

Respuesta :

mavila18

Answer:

[tex]E=3.69*10^{-11}\frac{V}{m}[/tex]

Explanation:

To solve this problem you use the following formula, for the calculation of the electric field along the axis of the dipole.

[tex]E=\frac{p}{2\pi \epsilon_ox^3}[/tex]   (1)

p: dipole moment = 6.16*10^-30 Cm

x: distance to the center of mass of the dipole = 3.00*10^-9m

eo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

You replace the values of the variables in the equation (1):

[tex]E=\frac{6.16*10^{-30}Cm}{2\pi(8.85*10^{-12}C^2/Nm^2)(3.00*10^{-9}m)^3}\\\\E=3.69*10^{-11}\frac{V}{m}[/tex]

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