Consider a dipole with charge q = 1.65 nC and separation d = 0.549 m. What is the potential a distance x = 0.699 m from the center of this dipole at an angle θ = 63.7° with respect to the dipole axis, as shown in the figure? V
Question
Consider a dipole with charge q = 1.65 nC and separation d = 0.549 m. What is the potential a distance x = 0.699 m from the center of this dipole at an angle θ = 63.7° with respect to the dipole axis, as shown in the figure? V
Solution
The potential V at a point distant r from the center of an electric dipole, at an angle θ to the dipole axis, is given by the formula:
V = k * p * cos(θ) / r^2
where:
- k is Coulomb's constant (8.99 * 10^9 N m^2/C^2),
- p is the dipole moment (q * d), and
- r is the distance from the center of the dipole.
First, calculate the dipole moment p:
p = q * d p = 1.65 * 10^-9 C * 0.549 m p = 9.04 * 10^-10 C*m
Then, calculate the potential V:
V = k * p * cos(θ) / r^2 V = 8.99 * 10^9 N m^2/C^2 * 9.04 * 10^-10 Cm * cos(63.7°) / (0.699 m)^2 V = 8.99 * 10^9 N m^2/C^2 * 9.04 * 10^-10 Cm * 0.443 / 0.48841 m^2 V = 3.6 * 10^3 V
So, the potential at a distance x = 0.699 m from the center of this dipole at an angle θ = 63.7° with respect to the dipole axis is approximately 3.6 kV.
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