An electric dipole consists of charges of ±6.00 µC that are 10.0 cm apart, as shown in the figure. Find the magnitude and direction of the electric field this dipole produces at point P, which is 7.00 cm from each charge. (k = 1/4pe0 = 8.99 × 10^9 N m²/C²)
The correct answer and explanation is :
To solve this problem, we need to calculate the electric field produced by the electric dipole at point P, which is located 7.00 cm from each charge. The electric dipole consists of two charges, ( +6.00 \, \mu C ) and ( -6.00 \, \mu C ), separated by a distance of 10.0 cm.
Step 1: Understanding the Problem
The dipole consists of two equal but opposite charges placed a certain distance apart. The electric field at point P depends on the positions of these charges relative to point P and their individual electric fields. The distance of 7.00 cm from each charge suggests that point P is located on the perpendicular bisector of the dipole. The electric fields from the two charges will combine vectorially to give the total electric field at point P.
Step 2: Electric Field Due to a Point Charge
The electric field due to a point charge ( q ) at a distance ( r ) is given by the formula:
[
E = \frac{k |q|}{r^2}
]
Where:
- ( k ) is Coulomb’s constant ( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) ),
- ( q ) is the charge (in Coulombs),
- ( r ) is the distance from the charge (in meters).
Step 3: Electric Field from Each Charge
For each charge, we can calculate the electric field at point P. The distance from each charge to point P is ( r = 7.00 \, \text{cm} = 0.07 \, \text{m} ).
For the positive charge ( q = +6.00 \, \mu C = +6.00 \times 10^{-6} \, \text{C} ):
[
E_+ = \frac{(8.99 \times 10^9) \times (6.00 \times 10^{-6})}{(0.07)^2} = 1.53 \times 10^6 \, \text{N/C}
]
For the negative charge ( q = -6.00 \, \mu C = -6.00 \times 10^{-6} \, \text{C} ):
[
E_- = \frac{(8.99 \times 10^9) \times (6.00 \times 10^{-6})}{(0.07)^2} = 1.53 \times 10^6 \, \text{N/C}
]
Step 4: Direction of Electric Fields
- The electric field due to the positive charge ( E_+ ) points away from the positive charge, so it points toward the right at point P.
- The electric field due to the negative charge ( E_- ) points toward the negative charge, so it points to the left at point P.
Step 5: Resultant Electric Field
Since the fields are of equal magnitude but opposite in direction, they will cancel each other out along the line connecting the charges. However, the components of the fields perpendicular to this line (along the perpendicular bisector) will add up. The total electric field at point P can be calculated by considering only the perpendicular components of the electric fields from each charge.
Using the symmetry of the problem, the total electric field at point P will be twice the component of one field perpendicular to the axis of the dipole:
[
E_{\text{total}} = 2 E \sin(\theta)
]
Where ( \theta = 90^\circ ) because point P lies on the perpendicular bisector of the dipole. Thus:
[
E_{\text{total}} = 2 \times 1.53 \times 10^6 = 3.06 \times 10^6 \, \text{N/C}
]
Final Answer:
The magnitude of the electric field at point P is approximately ( 3.06 \times 10^6 \, \text{N/C} ). The direction of the electric field is along the perpendicular bisector of the dipole, pointing from the positive charge to the negative charge.