An electric dipole is made of two charges of equal magnitudes and opposite signs. The
positive charge, q = 1.0μC, is located at the point (x, y, z) = (0.00cm, 1.0cm, 0.00cm), while
the negative charge is located at the point (x, y, z) = (0.00cm,−1.0cm, 0.00cm). How much
work will be done by an electric field ~E
= (3.0 × 106N/C)~i
to bring the dipole to its stable
equilibrium position?
To find the work done by an electric field to bring the dipole to its stable equilibrium position, we first need to determine the potential energy of the dipole at both the initial and final positions.
The potential energy of an electric dipole in an external electric field is given by the formula:
U = -p * E * cosθ
Where:
- U is the potential energy of the dipole
- p is the electric dipole moment (product of the magnitude of the charge and the separation distance between the charges)
- E is the magnitude of the electric field
- θ is the angle between the electric dipole moment and the electric field
In this case, the dipole is brought to its stable equilibrium position, which means both charges align with the electric field. Therefore, the angle θ will be 0 degrees, and cosθ will be equal to 1.
Let's calculate the potential energy at the initial position:
- p = q * d, where d is the separation distance between the charges and q is the charge. In this case, d = 2 cm.
- p = (1.0 μC) * (2 cm) = 2 μC.cm
- E = 3.0 × 10^6 N/C
U_initial = -p * E * cosθ
U_initial = -(2 μC.cm) * (3.0 × 10^6 N/C) * 1
U_initial = -6 × 10^6 μNm
Now let's calculate the potential energy at the final position (stable equilibrium):
Since both charges align with the electric field, the dipole moment and the electric field will be parallel, and cosθ will be equal to 1.
U_final = -p * E * cosθ
U_final = -(2 μC.cm) * (3.0 × 10^6 N/C) * 1
U_final = -6 × 10^6 μNm
The work done by the electric field to bring the dipole to its stable equilibrium position is the change in potential energy:
Work = U_final - U_initial
Work = (-6 × 10^6 μNm) - (-6 × 10^6 μNm)
Work = 0
Therefore, the work done by the electric field to bring the dipole to its stable equilibrium position is zero.
To calculate the work done by an electric field to bring the dipole to its stable equilibrium position, we need to find the electric potential energy of the dipole in its initial and final positions.
The electric potential energy of a dipole in an electric field is given by the formula:
U = -p · E
where U is the electric potential energy, p is the dipole moment, and E is the electric field vector.
The dipole moment is given by the formula:
p = q · d
where q is the magnitude of the charge and d is the distance between the charges.
Given:
q = 1.0 μC = 1.0 × 10^-6 C (charge magnitude)
E = 3.0 × 10^6 N/C (electric field magnitude)
d = 2.0 cm = 0.02 m (distance between charges)
First, calculate the dipole moment:
p = q · d
p = (1.0 × 10^-6 C) · (0.02 m)
p = 2.0 × 10^-8 C · m
Next, calculate the work done by the electric field:
U = -p · E
U = -(2.0 × 10^-8 C · m) · (3.0 × 10^6 N/C)
U = -6.0 × 10^-2 J
Therefore, the work done by the electric field to bring the dipole to its stable equilibrium position is -6.0 × 10^-2 J.