A long, thin rod (length = 4.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a +5.0 µC point charge is fixed to one end of the rod, while a -5.0 µC point charge is fixed to the other end. Everywhere in the x, y plane there is a constant external electric field (magnitude = 5.00 103 N/C) that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.
force on each = Q E
= 5 *10^-6 * 5 * 10^3 = 25 * 10^-3 N
torque resulting from each = F D
= 25 * 10^-3 * 2 = 50 * 10^-3 Nm
both give same sign torque so total torque = 100 *10^-3 Nm
= 0.1 Nm
To find the magnitude of the net torque applied to the rod with respect to the z-axis, we can use the formula:
τ = r × F
where τ is the torque, r is the position vector from the axis of rotation to the point of application of the force, and F is the force vector.
In this case, the net torque applied to the rod is due to two forces: the force exerted by the external electric field and the force exerted by the electric field between the two point charges on the ends of the rod.
Let's consider each force separately.
1. Torque due to the force exerted by the external electric field:
Since the external electric field is perpendicular to the rod and the axis of rotation, the torque due to this force is zero. This is because the position vector (r) and the force vector (F) are parallel to each other, yielding a cross product of zero.
2. Torque due to the force exerted by the electric field between the two point charges:
The force (F) between two point charges is given by Coulomb's Law:
F = k * q1 * q2 / r^2
where k is the electrostatic constant (9.0 × 10^9 Nm^2/C^2), q1 and q2 are the charges (+5.0 µC = 5.0 × 10^-6 C and -5.0 µC = -5.0 × 10^-6 C), and r is the distance between the charges (4.0 m).
Let's calculate the force (F) between the two point charges:
F = (9.0 × 10^9 Nm^2/C^2) * (5.0 × 10^-6 C) * (-5.0 × 10^-6 C) / (4.0 m)^2
F = -56.25 N
To calculate the torque due to this force, we need to determine the position vector (r). Since the midpoint of the rod is at the origin, the position vector for both point charges is half the length of the rod (2.0 m). Note that the torque will have opposite directions for the two charges.
Now, we can calculate the torque for each charge using the formula:
τ = r × F
For the positive charge at one end of the rod:
τ1 = (2.0 m) * (-56.25 N) = -112.5 Nm
For the negative charge at the other end of the rod:
τ2 = (-2.0 m) * (-56.25 N) = 112.5 Nm
The net torque is the sum of the torques produced by the two charges:
τ_net = τ1 + τ2 = -112.5 Nm + 112.5 Nm = 0 Nm
Therefore, the magnitude of the net torque applied to the rod with respect to the z-axis is 0 Nm.
To find the magnitude of the net torque applied to the rod, we can use the following formula:
τ = r × F
where τ is the torque, r is the position vector, and F is the force.
In this case, since the electric field is perpendicular to the rod, the force acting on each charge can be calculated using the formula:
F = q * E
where q is the charge and E is the electric field.
Let's first calculate the individual torques on both charges and then find the net torque.
For the positive charge (+5.0 µC):
The force acting on the positive charge is given by:
F1 = q1 * E
= (+5.0 µC) * (5.00 × 10^3 N/C) [Substituting the values]
= 25 × 10^-6 C * 5.00 × 10^3 N/C
= 125 × 10^-3 N
= 0.125 N
The position vector r1 for the positive charge is half the length of the rod, in the negative x direction because the charge is fixed at one end of the rod. Therefore, r1 = -2.0 m î
τ1 = r1 × F1
= (-2.0 m î) × (0.125 N ĵ)
= -0.25 N m k̂
For the negative charge (-5.0 µC):
The force acting on the negative charge is given by:
F2 = q2 * E
= (-5.0 µC) * (5.00 × 10^3 N/C) [Substituting the values]
= -25 × 10^-6 C * 5.00 × 10^3 N/C
= -125 × 10^-3 N
= -0.125 N
The position vector r2 for the negative charge is also half the length of the rod, in the positive x direction because the charge is fixed at the other end of the rod. Therefore, r2 = 2.0 m î
τ2 = r2 × F2
= (2.0 m î) × (-0.125 N ĵ)
= -0.25 N m k̂
Now, to find the net torque, we can add the individual torques:
τ_net = τ1 + τ2
= -0.25 N m k̂ + (-0.25 N m k̂)
= -0.5 N m k̂
The magnitude of the net torque is the absolute value of τ_net:
Magnitude of net torque = |τ_net|
= |-0.5 N m k̂|
= 0.5 N m
Therefore, the magnitude of the net torque applied to the rod with respect to the z-axis is 0.5 N m.