Two point charges are separated by 25.0cm (see the figure(Figure 1) ). Assume that q1 = 6.05nC and q2 = 10.5nC .
Find the magnitude of the net electric field these charges produce at point A. In N/C
What would be the magnitude of the electric force this combination of charges would produce on a proton at A? In N
Show transcribed image text Two point charges are separated by 25.0cm (see the figure(Figure 1) ). Assume that q1 = 6.05nC and q2 = 10.5nC . Find the magnitude of the net electric field these charges produce at point A. In N/C What would be the magnitude of the electric force this combination of charges would produce on a proton at A? In N
The Correct Answer and Explanation is :
To solve the problem, we need to calculate two quantities:
- The net electric field at point A due to the two charges (q1 and q2).
- The electric force on a proton at point A due to the net electric field.
Given:
- Distance between the charges: ( d = 25.0 \, \text{cm} = 0.25 \, \text{m} )
- ( q_1 = 6.05 \, \text{nC} = 6.05 \times 10^{-9} \, \text{C} )
- ( q_2 = 10.5 \, \text{nC} = 10.5 \times 10^{-9} \, \text{C} )
- The charge of a proton: ( q_p = 1.6 \times 10^{-19} \, \text{C} )
Electric Field Due to a Point Charge:
The electric field ( E ) at a distance ( r ) from a point charge ( q ) is given by Coulomb’s Law:
[
E = \frac{k |q|}{r^2}
]
Where:
- ( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 ) is Coulomb’s constant.
- ( r ) is the distance from the charge to the point where the field is calculated.
Step 1: Calculate the Electric Fields Due to Each Charge
Assuming that the point A is equidistant from both charges, the distances between ( q_1 ) and A, and between ( q_2 ) and A, will both be ( r = 0.25 \, \text{m} ).
For charge ( q_1 ):
[
E_1 = \frac{k |q_1|}{r^2} = \frac{8.99 \times 10^9 \times 6.05 \times 10^{-9}}{(0.25)^2} \approx 1.73 \times 10^3 \, \text{N/C}
]
For charge ( q_2 ):
[
E_2 = \frac{k |q_2|}{r^2} = \frac{8.99 \times 10^9 \times 10.5 \times 10^{-9}}{(0.25)^2} \approx 3.79 \times 10^3 \, \text{N/C}
]
Step 2: Find the Direction of the Electric Fields
- If both charges are of the same sign, the electric fields at point A will be in opposite directions (since they repel each other).
- If the charges are of opposite sign, the electric fields will point towards each other (since opposite charges attract).
Step 3: Net Electric Field at Point A
If ( q_1 ) and ( q_2 ) are of opposite signs, the magnitudes of the electric fields at point A will add up:
[
E_{\text{net}} = E_1 + E_2 = 1.73 \times 10^3 + 3.79 \times 10^3 = 5.52 \times 10^3 \, \text{N/C}
]
Step 4: Calculate the Force on a Proton at Point A
The force on a proton at point A is given by:
[
F = q_p E_{\text{net}}
]
Substitute the values:
[
F = 1.6 \times 10^{-19} \times 5.52 \times 10^3 \approx 8.83 \times 10^{-16} \, \text{N}
]
Final Answer:
- The magnitude of the net electric field at point A is ( \mathbf{5.52 \times 10^3 \, \text{N/C}} ).
- The magnitude of the electric force on a proton at point A is ( \mathbf{8.83 \times 10^{-16} \, \text{N}} ).
Explanation:
This problem combines Coulomb’s Law for calculating electric fields and forces. First, we calculated the electric field due to each charge separately. Then, we found the net electric field by considering the direction of the fields based on the sign of the charges. Finally, we used the net electric field to find the force on a proton at point A.