An electric dipole consists of 2.0 g spheres charged to 5.0 nC (positive and negative) at the ends of a 12 cm long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength 1400V, then released. What is the dipole’s angular velocity at the instant it is aligned with the electric field?
The torque acting on the dipole in electric field of strength E= 1400 V/m (!!!!- units) is M=F•(d/2), where d=12 cm and F=qE.
From mechanics (Newton’s second law for rotation)
M=I•ε,
where I=(md^2)/12 is the moment of inertia of the rod about the axis passing through the center, and ε is the angular acceleration of the rod.
Therefore,
qE •(d/2) = ε• (md^2)/12,
ε=6qE/md=0.175 rad/s.
Since the angular velocity is ω= εt and angular displacement is φ= ε t^2/2,
ω=sqroot(2 ε•φ)= sqroot(2 0.175•π/2)=0.74 rad/s.
Why did the electric dipole start a dating app? Because it was looking for a strong "field"ship! *ba dum tss*
Now, back to your question. To calculate the dipole's angular velocity, we can use the formula:
Angular velocity (ω) = Torque (τ) / Moment of inertia (I)
The torque experienced by the dipole can be calculated using the formula:
Torque (τ) = electric field (E) * dipole moment (p) * sin(θ),
where E is the electric field strength and θ is the angle between the dipole moment and the electric field.
In this case, when the dipole is aligned with the electric field, θ = 0, and sin(θ) = 0. Therefore, the torque becomes zero.
Since there is no torque acting on the system, the angular velocity will also be zero. So, the dipole's angular velocity at the instant it aligns with the electric field is 0.
To find the dipole's angular velocity at the instant it is aligned with the electric field, we can use the principle of conservation of energy.
The potential energy of the dipole in an electric field is given by:
U = -pEcosθ,
where U is the potential energy, p is the electric dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field.
When the dipole is aligned with the electric field (θ = 0°), the potential energy is minimum. At this point, all the potential energy is converted into kinetic energy.
The kinetic energy of the rotating dipole is given by:
K = (1/2)Iω^2,
where K is the kinetic energy, I is the moment of inertia of the dipole, and ω is the angular velocity.
Since there is no external torque acting on the dipole (due to the frictionless pivot), the angular momentum is conserved:
L = Iω = constant.
At the instant when the dipole is aligned with the electric field, the initial potential energy U is converted to the final kinetic energy K.
Therefore, -pE = (1/2)Iω^2.
We can find the moment of inertia of the dipole using the formula:
I = 2/3 mL^2,
where m is the mass of each sphere, and L is the length of the rod.
Given data:
m = 2.0 g = 0.002 kg (mass of each sphere)
L = 12 cm = 0.12 m (length of the rod)
p = 5.0 nC = 5.0 × 10^-9 C (electric dipole moment)
E = 1400 V/m (electric field strength)
Substituting these values into the formula for the moment of inertia:
I = (2/3)(0.002 kg)(0.12 m)^2,
I = 0.000048 kg*m^2.
Substituting the known values into the equation -pE = (1/2)Iω^2:
-(5.0 × 10^-9 C)(1400 V/m) = (1/2)(0.000048 kg*m^2)ω^2.
Simplifying the equation:
-(7.0 × 10^-6) = (2.4 × 10^-5)ω^2.
Solving for ω:
ω^2 = -(7.0 × 10^-6) / (2.4 × 10^-5),
ω^2 ≈ -0.292.
Since angular velocity cannot be negative, we can take the square root of the positive value:
ω ≈ √(0.292) = 0.54 rad/s.
Therefore, the dipole's angular velocity at the instant it is aligned with the electric field is approximately 0.54 rad/s.
To find the dipole's angular velocity at the instant it is aligned with the electric field, we can use the principles of torque and rotational motion.
First, we need to find the net torque acting on the dipole when it is aligned with the electric field. The torque experienced by an electric dipole in an electric field can be given by the equation:
Torque = Electric dipole moment x Electric field x sin(θ)
where:
- Electric dipole moment (p) is equal to the charge (q) multiplied by the distance (d) between the charges: p = qd
- Electric field (E) is given as 1400V in this case.
- θ is the angle between the electric field and the dipole moment, which is 0 degrees when the dipole is aligned with the electric field.
Now let's calculate the electric dipole moment:
- The charge (q) is given as 5.0 nC, which is equal to 5.0 x 10^(-9) C.
- The distance (d) between the charges is given as 12 cm, which is equal to 0.12 m.
Therefore, the electric dipole moment (p) is given by:
p = (5.0 x 10^(-9) C) x (0.12 m) = 6.0 x 10^(-10) C⋅m
Now let's calculate the torque:
Torque = (6.0 x 10^(-10) C⋅m) x (1400 V) x sin(0°)
= (6.0 x 10^(-10) C⋅m) x (1400 V) x sin(0)
= 0
Since the angle θ is 0 degrees when the dipole is aligned with the electric field, the sin(0) is also 0. Therefore, the torque is 0 when the dipole is aligned with the electric field.
Now, we can apply the principle of conservation of angular momentum. When the dipole is released, it will start rotating due to the torque acting on it until it aligns with the electric field. The angular momentum (L) of the system is conserved, and it can be expressed as:
L = Moment of inertia x Angular velocity
At the initial and final positions (aligned with the electric field), the angular momentum is the same. Since the moment of inertia remains constant, the angular velocity will be the same in both positions.
Therefore, the dipole's angular velocity at the instant it is aligned with the electric field is zero.