Two point charges are separated by 25.0cm. Assume that q1 = -6.25 nC and q2 = -12.5nC .
Find the magnitude of the net electric field these charges produce at (a)point A. (b) point B.
(c) What would be the magnitude and direction of the electric force this combination of charges produce on a proton at A?
The Correct Answer and Explanation is :
Let’s break down the problem step-by-step. For this, we will apply the formula for the electric field produced by a point charge:
[
E = k_e \cdot \frac{|q|}{r^2}
]
Where:
- (E) is the electric field,
- (k_e) is Coulomb’s constant, (8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2),
- (q) is the charge,
- (r) is the distance from the charge to the point where the electric field is being measured.
Part (a) – Net Electric Field at Point A
The electric field due to a point charge points away from the charge if it is positive, and toward the charge if it is negative.
- Distance from (q_1) to point A: Since (q_1) is placed on the left, the distance from (q_1) to point A is 10 cm (0.1 m).
- Distance from (q_2) to point A: Point A is placed 15 cm (0.15 m) from (q_2).
Now, calculate the individual electric fields:
- Electric field due to (q_1) at point A:
[
E_1 = k_e \cdot \frac{|q_1|}{r_1^2} = 8.99 \times 10^9 \times \frac{6.25 \times 10^{-9}}{(0.1)^2}
]
- Electric field due to (q_2) at point A:
[
E_2 = k_e \cdot \frac{|q_2|}{r_2^2} = 8.99 \times 10^9 \times \frac{12.5 \times 10^{-9}}{(0.15)^2}
]
Both fields point toward their respective charges, and we need to sum them to find the net electric field at A.
Part (b) – Electric Field at Point B
Point B is 15 cm (0.15 m) from (q_1) and 10 cm (0.1 m) from (q_2). Using the same method as in part (a), calculate the individual electric fields at point B and find the net field.
Part (c) – Electric Force on a Proton at A
The force on a proton is given by:
[
F = q_{\text{proton}} \cdot E_{\text{net}}
]
Where:
- (q_{\text{proton}}) is the charge of the proton, (1.6 \times 10^{-19} \, \text{C}),
- (E_{\text{net}}) is the net electric field at point A (from part (a)).
The direction of the force will be the same as the direction of the electric field at point A because the proton is positively charged.
Let me compute these values and generate the image.
Here are the results:
Part (a) – Net Electric Field at Point A:
The net electric field at point A is approximately ( 10613.2 \, \text{N/C} ).
Part (b) – Net Electric Field at Point B:
The net electric field at point B is approximately ( 13734.7 \, \text{N/C} ).
Part (c) – Electric Force on a Proton at Point A:
The electric force on a proton at point A is approximately ( 1.70 \times 10^{-15} \, \text{N} ), and it will be directed toward the left, since the electric field points toward the negative charges.
Now, I will generate the image based on the provided diagram.
Here is the diagram showing the electric fields at points A and B, as well as the electric force on a proton at point A. Let me know if you need any further clarification or help!
