Two point charges are separated by 25.0 cm Assume that q1 = 6.25 nC and q2 = 12.5 nC . A) Find the magnitude of the net electric field these charges produce at point A. (N/C) B) Find the direction of the net electric field these charges produce at point A. (right or left) C) Find the magnitude of the net electric field these charges produce at point B. (N/C) D) Find the direction of the net electric field these charges produce at point B. (right or left) E) What would be the magnitude of the electric force this combination of charges would produce on a proton at A? (N) F) What would be the direction of the electric force this combination of charges would produce on a proton at A? (right or left
The Correct Answer and Explanation is:
To solve the problems involving electric fields due to two point charges, we can use Coulomb’s Law and the principle of superposition. Here’s the step-by-step approach to each part:
Coulomb’s Law:
The electric field (EEE) due to a point charge is given by the formula: E=k∣q∣r2E = \frac{k |q|}{r^2}E=r2k∣q∣
Where:
- EEE is the electric field,
- kkk is Coulomb’s constant (8.99×109 N\cdotpm2/C28.99 \times 10^9 \, \text{N·m}^2/\text{C}^28.99×109N\cdotpm2/C2),
- qqq is the charge producing the field,
- rrr is the distance from the charge to the point where the electric field is measured.
Also, the direction of the electric field:
- Away from the charge if the charge is positive.
- Toward the charge if the charge is negative.
Assumptions:
- q1=6.25 nC=6.25×10−9 Cq_1 = 6.25 \, \text{nC} = 6.25 \times 10^{-9} \, \text{C}q1=6.25nC=6.25×10−9C,
- q2=12.5 nC=12.5×10−9 Cq_2 = 12.5 \, \text{nC} = 12.5 \times 10^{-9} \, \text{C}q2=12.5nC=12.5×10−9C,
- The separation distance between q1q_1q1 and q2q_2q2 is 25.0 cm (r=0.25 mr = 0.25 \, \text{m}r=0.25m),
- The proton has a charge of 1.6×10−19 C1.6 \times 10^{-19} \, \text{C}1.6×10−19C and will feel a force due to the electric field.
A) Net Electric Field at Point A:
The electric field at point A will be the vector sum of the fields due to q1q_1q1 and q2q_2q2. The magnitude of the field from each charge can be calculated using the formula above.
Field due to q1q_1q1 at A:
E1=k∣q1∣r12E_1 = \frac{k |q_1|}{r_1^2}E1=r12k∣q1∣
Field due to q2q_2q2 at A:
E2=k∣q2∣r22E_2 = \frac{k |q_2|}{r_2^2}E2=r22k∣q2∣
After finding the fields, sum them, keeping in mind the direction of each field.
B) Direction of the Electric Field at Point A:
- The direction of the electric field depends on the sign of the charges. Since both charges are positive, both fields will point away from the charges.
- The direction at point A will be determined by the relative positions of the charges and point A.
C) Net Electric Field at Point B:
The procedure is similar to part A. Calculate the fields from both charges at point B and then find the net electric field.
D) Direction of the Electric Field at Point B:
Determine the direction of each field at point B, considering the same principles as in part B.
E) Magnitude of the Electric Force on a Proton at A:
The force on the proton is given by: F=qproton×EnetF = q_{\text{proton}} \times E_{\text{net}}F=qproton×Enet
Where EnetE_{\text{net}}Enet is the net electric field at point A. Use the magnitude of the electric field calculated in part A.
F) Direction of the Electric Force on a Proton at A:
- Since the proton has a positive charge, the force will be in the direction of the net electric field at point A (since positive charge moves in the direction of the electric field).
