The dipole moment of a water molecule is 6.37E-30 Cm. What is the electric potential's magnitude 1.58 nm from a water molecule along the axis of the dipole?
To find the electric potential's magnitude at a distance of 1.58 nm from a water molecule along the axis of its dipole, we can use the formula for the electric potential due to a dipole:
V = k * (p / r^2)
Where:
V is the electric potential due to the dipole
k is the Coulomb constant (9.0 x 10^9 Nm^2/C^2)
p is the dipole moment of the water molecule (6.37 x 10^-30 Cm)
r is the distance from the water molecule's dipole along the axis (1.58 nm = 1.58 x 10^-9 m)
Plugging in the values, we get:
V = (9.0 x 10^9 Nm^2/C^2) * (6.37 x 10^-30 Cm) / (1.58 x 10^-9 m)^2
Now, let's calculate the value using a calculator:
V = (9.0 x 10^9) * (6.37 x 10^-30) / (1.58 x 10^-9)^2
V ≈ 2.439 x 10^-20 Volts
Therefore, the magnitude of the electric potential at a distance of 1.58 nm from a water molecule along the axis of its dipole is approximately 2.439 x 10^-20 Volts.