A charge, q1, of +4.4 micro-coulombs and another unknown charge, q2, are surrounded by a surface. The total flux through the surface is -9.2 x 106 (N m2)/C. These charges are now placed on the x axis. q1 is placed at x = 0, and q2 is placed at x = +114.51 cm. What is the magnitude of the total electric field produced by both charges exactly halfway between the two charges? Answer in MN/C (divide answer by 1x106 to convert to MN/C).
To find the magnitude of the total electric field at the midpoint between the charges, we can use the principle of superposition. The electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge.
Let's calculate the electric field produced by each charge at the midpoint and then sum them to find the total electric field.
1. Electric field due to q1 at the midpoint:
Since q1 is located at x = 0, the distance from q1 to the midpoint, let's call it r₁, is half the distance between the charges (57.255 cm or 0.57255 m).
To calculate the electric field, we will use Coulomb's Law: E = kq / r², where k is the electrostatic constant (k = 9 x 10^9 N m²/C²).
E₁ = (k * q₁) / r₁²
= (9 x 10^9 N m²/C²) * (4.4 x 10^-6 C) / (0.57255 m)²
2. Electric field due to q₂ at the midpoint:
Since q₂ is located at x = +114.51 cm or 1.1451 m, the distance from q₂ to the midpoint, let's call it r₂, is also half the distance between the charges (0.57255 m).
E₂ = (k * q₂) / r₂²
= (9 x 10^9 N m²/C²) * (q₂) / (0.57255 m)²
3. Total electric field at the midpoint:
The total electric field is the vector sum of the individual electric fields at the midpoint.
E_total = E₁ + E₂
Finally, divide the answer by 1 x 10^6 to convert the units from N/C to MN/C.
Please provide the value of q₂ in order to calculate the total electric field at the midpoint.