Question

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?

Answer 1

The magnitude of the electric field created by a water molecule at a point along the x-axis is calculated using the dipole electric field formula. Given the dipole moment and distance, the electric field's magnitude at the specified distance is approximately 97.2 N/C.

Step-by-step explanation:

To calculate the magnitude of the electric field created by a water molecule at a certain point along the x-axis, we can use the formula for the electric field due to a dipole at points along the axis of the dipole:

E = (1 / (4πε0)) * (2 * p) / x3

where ε0 is the vacuum permittivity (ε0 = 8.85 × 10-12 C2/N·m2), p is the dipole moment, and x is the distance from the dipole.

Given:

- Dipole moment (p) = 6.16 × 10-30 C·m

- Distance from the dipole (x) = 3.00 × 10-9 m

Plugging these values into the formula gives us:

E = (1 / (4π * 8.85 × 10-12)) * (2 * 6.16 × 10-30) / (3.00 × 10-9)3

Calculating this we get:

E = 9.72 × 101 N/C

The magnitude of the electric field at x = 3.00 × 10-9 m is approximately 97.2 N/C.

Answer 2

`E=3.69*10^(-11)(V)/(m)`

Step-by-step explanation:

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

`E=(p)/(2\pi \epsilon_ox^3)` (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):

`E=(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)(V)/(m)`